problem solving multiplying polynomials

Multiplying Polynomials Word Problems - Examples & Practice - Expii

Multiplying polynomials word problems - examples & practice, explanations (3).

Polynomial Word Problems

Image by geralt via Pixabay ( CC0 )

As of right now, you should know how to add , subtract , and multiply polynomials.

So when we look at word problems they'll only be asking you to add, subtract, or multiply polynomials.

The trickier part is figuring out which they want you to use. Fortunately, the problems usually deal with shapes which can help clue us in on what to use. If it references perimeter, chances are you need to use addition or subtraction. If references area or volume, chances are you need to use multiplication.

Let's look at an example.

Linda has a rectangular garden that's 3x−5 feet by 2x+8 feet. She wants to put a fence around the perimeter of the garden. If each foot of fence costs 5 dollars, how much will the total fencing around the garden cost?

Step 1: Draw a Diagram

Seriously, don't skip this step. Diagrams are a great way to help visualize the problem and keep things in order. Plus you get to make a quick doodle.

Image Source: Expii

Here's Linda's garden. Each side is either 3x−5 feet or 2x+8 feet.

Step 2: Figure out What to Use

We want to find the price of the fencing that will surround the garden. To do this, we first need to figure out how much fencing there is.

This is a perimeter problem. Which means, we'll need to add up the sides.

Step 3: Set up the Problem

The last trickier step, is to set the problem up . First, we need to find the perimeter of the rectangle. perimeter=(3x−5)+(2x+8)+(3x−5)+(2x+8) Then, we multiply this by the price per foot of fencing, which is 5 dollars. 5(perimeter)=5[(3x−5)+(2x+8)+(3x−5)+(2x+8)]

Step 4: Solve

In my opinion, this is the easiest step of word problems. We already have it all set up, and finally we just add together some polynomials which we already know how to do. 5[(3x−5)+(2x+8)+(3x−5)+(2x+8)]=5[3x−5+2x+8+3x−5+2x+8]=5[(3x+2x+3x+2x)+(−5+8−5+8)]=5[10x+6]=50x+30 It will cost 50x+30 dollars to put fencing around this garden.

Related Lessons

Word problems involving binomials are often related to basic geometry (like the concept of area , for example).

An important thing to remember is that word problems are just presenting you with familiar information in a new way (i.e., in words). You already know how to handle these problems , so don't let the fact that they're word problems throw you off.

Let's look at an example:

Image source: by Hannah Bonville

First, let's remind ourselves of how to find area. What's the formula for finding the area of the shape above?

Area=length+width

Area=length×width

Area=(2×length)+(2×width)

Area=length/width

(Video) Polynomials 06 Multiply Polynomials Word Problem

by Mister Zuidema

This video by Mister Zuidema walks through a word problem with polynomials.

Let's look at it.

Sally has a photo which is 20 cm tall and 32 cm wide. She wants to put a frame around the photo which is the same width all around. What is the expression for the area of the framed photo.

First, we draw out the diagram of the photo and the frame. We don't know the width of the frame, so we label it x. We want the area of the whole frame, so we need to write an expression that is equal to the length times the height of the frame. We know the photo is 32 cm wide and the frame is x wide on each side. So, the length of the frame is (32+2x). We find the height of the frame similarly. We know the height of the photo is 20 and each side of the frame is x, so the height is (20+2x). Now, the area is the length times the width, so we multiply binomials. In the video, he uses a different way of multiplying, but I will be using FOIL . We see,

(32+2x)×(20+2x)=(32)(20)+(32)(2x)+(2x)(20)+(2x)(2x)=640+64x+40x+4x2=640+104x+4x2

The expression that represents the area of the whole frame is 640+104x+4x2 which is equal to 4x2+104x+640

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5.4: Multiplying Polynomials

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Learning Objectives

• Multiply a polynomial by a monomial.
• Multiply a polynomial by a binomial.
• Multiply a polynomial by any size polynomial.
• Recognize and calculate special products.
• Multiply polynomial functions.

Multiplying by a Monomial

Recall the product rule for exponents: if \(m\) and \(n\) are positive integers, then

\[x^{m}\cdot x^{n}=x^{m+n}\]

In other words, when multiplying two expressions with the same base, add the exponents. This rule applies when multiplying a monomial by a monomial. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For example,

\(\begin{array} {cl} {3x\cdot 5x^{2} = 3\cdot 5\cdot x^{1}\cdot x^{2}}&{\color{Cerulean}{Commutative\: property}}\\{=15x^{1+2}}&{\color{Cerulean}{Product\:rule\:for\:exponents}}\\{=15x^{3}}&{} \end{array}\)

To multiply a polynomial by a monomial, apply the distributive property and then simplify each term.

Example \(\PageIndex{1}\)

\(−5x(4x−2)\).

In this case, multiply the monomial, \(−5x\), by the binomial, \(4x−2\). Apply the distributive property and then simplify.

\(-20x^{2}+10x\)

Example \(\PageIndex{2}\)

\(2x^{2}(3x^{2}−5x+1)\).

Apply the distributive property and then simplify.

\(6x^{4}-10x^{3}+2x^{2}\)

Example \(\PageIndex{3}\)

\(−3ab^{2}(a^{2}b^{3}+2a^{3}b−6ab−4)\).

To summarize, multiplying a polynomial by a monomial involves the distributive property and the product rule for exponents. Multiply all of the terms of the polynomial by the monomial. For each term, multiply the coefficients and add exponents of variables where the bases are the same.

Exercise \(\PageIndex{1}\)

\(−5x^{2}y(2xy^{2}−3xy+6x^{2}y−1)\).

\(−10x^{3}y^{3}+15x^{3}y^{2}−30x^{4}y^{2}+5x^{2}y\)

Multiplying by a Binomial

In the same way that we used the distributive property to find the product of a monomial and a binomial, we will use it to find the product of two binomials.

\[\begin{aligned} \color{Cerulean}{(a+b)}\color{black}{(c+d)} \\ &=\color{Cerulean}{(a+b)}\color{black}{\cdot c+}\color{Cerulean}{(a+b)}\color{black}{\cdot d} \\ &=ac+bc+ad+bd \\ &=ac+ad+bc+bd \end{aligned}\]

Here we apply the distributive property multiple times to produce the final result. This same result is obtained in one step if we apply the distributive property to \(a\) and \(b\) separately as follows:

This is often called the FOIL method. We add the products of the first terms of each binomial \(ac\), the \(o\)uter terms \(ad\), the \(i\)nner terms \(bc\), and finally the last terms \(bd\). This mnemonic device only works for products of binomials; hence it is best to just remember that the distributive property applies.

Example \(\PageIndex{4}\)

\((2x+3)(5x−2)\).

Distribute \(2x\) and then distribute \(3\).

Simplify by combining like terms.

\(=10x^{2}+11x-6\)

\(10x^{2}+11x-6\)

Example \(\PageIndex{5}\)

\((\frac{1}{2}x−\frac{1}{4})(\frac{1}{2}x+\frac{1}{4})\).

Distribute \(\frac{1}{2}x\) and then distribute \(−\frac{1}{4}\).

\(\begin{aligned} (\frac{1}{2}x−\frac{1}{4})(\frac{1}{2}x+\frac{1}{4}) &=\color{Cerulean}{\frac{1}{2}x}\color{black}{\frac{1}{2}x+}\color{Cerulean}{\frac{1}{2}x}\color{black}{\cdot\frac{1}{4}+}\color{OliveGreen}{\left( -\frac{1}{4} \right)}\color{black}{\cdot\frac{1}{2}x+}\color{OliveGreen}{\left(-\frac{1}{4} \right)}\color{black}{\cdot\frac{1}{4}} \\ &=\frac{1}{4}x^{2}+\frac{1}{8}x-\frac{1}{8}x-\frac{1}{16} \\ &=\frac{1}{4}x^{2}-\frac{1}{16} \end{aligned}\)

\(\frac{1}{4}x^{2}-\frac{1}{16}\)

Example \(\PageIndex{6}\)

\((3y^{2}−1)(2y+1)\).

\(6y^{3}+3y^{2}-2y-1\)

After applying the distributive property, combine any like terms.

Example \(\PageIndex{7}\)

\((x^{2}−5)(3x^{2}−2x+2)\).

After multiplying each term of the trinomial by \(x^{2}\) and \(−5\), simplify.

\(3x^{4}-2x^{3}-13x^{2}+10x-10\)

Example \(\PageIndex{8}\)

\((2x−1)^{3}\).

Perform one product at a time.

\(8x^{3}-12x^{2}+6x-1\)

At this point, it is worth pointing out a common mistake:

\((2x-1)^{3}\neq (2x)^{3}-(1)^{3}\)

The confusion comes from the product to a power rule of exponents, where we apply the power to all factors. Since there are two terms within the parentheses, that rule does not apply. Care should be taken to understand what is different in the following two examples:

\(\begin{aligned} (xy)^{2} &=x^{2}y^{2}\quad\color{Cerulean}{\checkmark} \\ (x+y)^{2} &\neq x^{2}+y^{2}\quad\color{red}{x} \end{aligned}\)

Exercise \(\PageIndex{2}\)

\((2x−3)(7x^{2}−5x+4)\).

\(14x^{3}-31x^{2}+23x-12\)

Product of Polynomials

When multiplying polynomials, we apply the distributive property many times. Multiply all of the terms of each polynomial and then combine like terms.

Example \(\PageIndex{9}\)

\((2x^{2}+x−3)(x^{2}−2x+5)\).

Multiply each term of the first trinomial by each term of the second trinomial and then combine like terms.

Aligning like terms in columns, as we have here, aids in the simplification process

\(2x^{4}-3x^{3}+5x^{2}+11x-15\)

Notice that when multiplying a trinomial by a trinomial, we obtain nine terms before simplifying. In fact, when multiplying an \(n\)-term polynomial by an m-term polynomial, we will obtain \(n × m\) terms. In the previous example, we were asked to multiply and found that

\((2x^{2}+x-3)(x^{2}-2x+5)=2x^{4}-3x^{3}+5x^{2}+11x-15\)

Because it is easy to make a small calculation error, it is a good practice to trace through the steps mentally to verify that the operations were performed correctly. Alternatively, we can check by evaluating any value for \(x\) in both expressions to verify that the results are the same. Here we choose \(x = 2\):

\(\begin{aligned} (2x^{2}+x-3)(x^{2}-2x+5)&=(2(\color{OliveGreen}{2}\color{black}{)^{2}+(}\color{OliveGreen}{2}\color{black}{)-3)((}\color{OliveGreen}{2}\color{black}{)^{2}-2(}\color{OliveGreen}{2}\color{black}{)+5)} \\ &=(8+2-3)(4-4+5) \\ &=(7)(5) \\ &=35 \end{aligned}\)

Because the results could coincidentally be the same, a check by evaluating does not necessarily prove that we have multiplied correctly. However, after verifying a few values, we can be fairly confident that the product is correct.

Exercise \(\PageIndex{3}\)

\((x^{2}−2x−3)^{2}\).

\(x^{4}−4x^{3}−2x^{2}+12x+9\)

Special Products

In this section, the goal is to recognize certain special products that occur often in our study of algebra. We will develop three formulas that will be very useful as we move along. The three should be memorized. We begin by considering the following two calculations:

\(\begin{array}{r|r} {(a+b)^{2}=(a+b)(a+b)}&{(a-b)^{2}=(a-b)(a-b)}\\{=a^{2}+ab+ba+b^{2}}&{=a^{2}-ab-ba+b^{2}}\\{=a^{2}+ab+ab+b^{2}}&{=a^{2}-ab-ab+b^{2}}\\{=a^{2}+2ab+b^{2}}&{=a^{2}-2ab+b^{2}} \end{array}\)

This leads us to two formulas that describe perfect square trinomials:

\[ (a+b)^{2}=a^{2}+2ab+b^{2} \]

\[ (a-b)^{2}=a^{2}-2ab+b^{2}\]

We can use these formulas to quickly square a binomial.

Example \(\PageIndex{10}\)

\((3x+5)^{2}\).

Here \(a=3x\) and \(b=5\). Apply the formula:

\(\begin{aligned} \color{Cerulean}{(a+b)^{2}} &\color{Cerulean}{ =\:\: a^{2}\:\:\:\:+2\:\:\:\:\:\:a\:\:\:\:\:b\:\:+\:\:b^{2}} \\ &\color{Cerulean}{\quad\:\:\: \downarrow\qquad\qquad\:\: \downarrow\:\:\:\:\: \downarrow\qquad\downarrow} \\ (3x+5)^{2}&=(3x)^{2}+2\cdot(3x)(5)+(5)^{2} \\ &=9x^{2}+30x+25\end{aligned}\)

\(9x^{2}+30x+25\)

This process should become routine enough to be performed mentally.

Example \(\PageIndex{11}\)

\((x−4)^{2}\).

Here \(a=x\) and \(b=4\). Apply the appropriate formula as follows:

\(\begin{aligned} \color{Cerulean}{(a-b)^{2}} &\color{Cerulean}{ =\:\: a^{2}\:\:\:\:-2\:\:\:\:a\:\:\:b\:\:+\:b^{2}} \\ &\color{Cerulean}{\quad\:\:\: \downarrow\qquad\quad\:\:\:\: \downarrow\:\:\: \downarrow\quad\:\:\:\downarrow} \\ (x-4)^{2}&=(x)^{2}-2\cdot(x)(4)+(4)^{2} \\ &=x^{2}-8x+16\end{aligned}\)

\(x^{2}-8x+16\)

Our third special product follows:

\(\begin{aligned}(a+b)(a-b)&=a^{2}-ab+ba-b^{2} \\ &=a^{2}\color{red}{-ab+ab}\color{black}{-b^{2}}\\&=a^{2}-b^{2} \end{aligned}\)

This product is called difference of squares:

\[(a+b)(a-b)=a^{2}-b^{2}\]

The binomials \((a+b)\) and \((a−b)\) are called conjugate binomials. Therefore, when conjugate binomials are multiplied, the middle term eliminates, and the product is itself a binomial.

Example \(\PageIndex{12}\)

\((7x+4)(7x−4)\).

\(49x^{2}-16\)

Exercise \(\PageIndex{4}\)

\((−5x+2)^{2}\).

\(25x^{2}−20x+4\)

Multiplying Polynomial Functions

We use function notation to indicate multiplication as follows:

Example \(\PageIndex{13}\)

\((f⋅g)(x)\), given \(f(x)=5x^{2}\) and \(g(x)=−x^{2}+2x−3\).

Multiply all terms of the trinomial by the monomial function \(f(x)\).

\(\begin{aligned} (f\cdot g)(x)&=f(x)\cdot g(x) \\ &=5x^{2}\cdot (-x^{2}+2x-3) \\ &=-5x^{4}+10x^{3}-15x^{2} \end{aligned}\)

\((f\cdot g)(x)=-5x^{4}+10x^{3}-15x^{2}\)

Example \(\PageIndex{14}\)

\((f⋅g)(−1)\), given \(f(x)=−x+3\) and \(g(x)=4x^{2}−3x+6\).

First, determine \((f⋅g)(x)\).

\(\begin{aligned} (f\cdot g)(x) &=f(x)\cdot g(x) \\ &=(-x+3)(4x^{2}-3x+6) \\ &=-4x^{3}+3x^{2}-6x+12x^{2}-9x+18 \\ &=-4x^{3}+15x^{2}-15x+18 \end{aligned}\)

\((f\cdot g)(x) = -4x^{3}+15x^{2}-15x+18

Next, substitute \(−1\) for the variable \(x\).

\(\begin{aligned} (f\cdot g)(\color{OliveGreen}{-1}\color{black}{)}&=-4(\color{OliveGreen}{-1}\color{black}{)^{3}+15(}\color{OliveGreen}{-1}\color{black}{)^{2}-15(}\color{OliveGreen}{-1}\color{black}{)+18} \\ &=-4\cdot (-1)+15\cdot 1+15+18 \\ &=4+15+15+18 \\ &=52 \end{aligned}\)

\((f\cdot g)(-1)=52\)

Because \((f⋅g)(−1)=f(−1)⋅g(−1)\), we could alternatively calculate \(f(−1)\) and \(g(−1)\) separately and then multiply the results (try this as an exercise). However, if we were asked to evaluate multiple values for the function \((f⋅g)(x)\), it would be best to first determine the general form, as we have in the previous example.

Key Takeaways

• To multiply a polynomial by a monomial, apply the distributive property and then simplify each of the resulting terms.
• To multiply polynomials, multiply each term in the first polynomial with each term in the second polynomial. Then combine like terms.
• The product of an \(n\)-term polynomial and an \(m\)-term polynomial results in an \(m × n\) term polynomial before like terms are combined.
• Check results by evaluating values in the original expression and in your answer to verify that the results are the same.
• Use the formulas for special products to quickly multiply binomials that occur often in algebra.

Exercise \(\PageIndex{5}\) Product of a Monomial and a Polynomial

• \(5x(−3x^{2}y)\)
• \((−2x^{3}y^{2})(−3xy^{4})\)
• \(\frac{1}{2}(4x−3)\)
• \(−\frac{3}{4}(\frac{2}{3}x−6)\)
• \(3x(5x−2)\)
• \(−4x(2x−1)\)
• \(x^{2}(3x+2)\)
• \(−6x^{2}(5x+3)\)
• \(2ab(4a−2b)\)
• \(5a^{2}b(a^{2}−b^{2})\)
• \(6x^{2}y^{3}(−3x^{3}y+xy^{2})\)
• \(3ab^{3}(−5ab^{3}+6a^{2}b)\)
• \(−\frac{1}{2}x^{2}y(4xy−10)\)
• \(−3x^{4}y^{2}(3x^{8}y^{3})\)
• \(2x^{2}(−5x^{3})(3x^{4})\)
• \(4ab(a^{2}b^{3}c)(a^{4}b^{2}c^{4})\)
• \(−2(5x^{2}−3x+4)\)
• \(\frac{4}{5}(25x^{2}−50xy+5y^{2})\)
• \(3x(5x^{2}−2x+3)\)
• \(−x(x^{2}+x−1)\)
• \(x^{2}(3x^{2}−5x−7)\)
• \(x^{3}(−4x^{2}−7x+9)\)
• \(\frac{1}{4}x^{4}(8x^{3}−2x^{2}+\frac{1}{2}x−5)\)
• \(−\frac{1}{3}x^{3}(\frac{3}{2}x^{5}−\frac{2}{3}x^{3}+\frac{9}{2}x−1)\)
• \(a^{2}b(a^{2}−3ab+b^{2})\)
• \(6a^{2}bc^{3}(2a−3b+c^{2})\)
• \(\frac{2}{3}xy^{2}(9x^{3}y−27xy+3xy^{3})\)
• \(−3x^{2}y^{2}(12x^{2}−10xy−6y^{2})\)
• Find the product of \(3x\) and \(2x^{2}−3x+5\).
• Find the product of \(−8y\) and \(y^{2}−2y+12\).
• Find the product of \(−4x\) and \(x^{4}−3x^{3}+2x^{2}−7x+8\).
• Find the product of \(3xy^{2}\) and \(−2x^{2}y+4xy−xy^{2}\).

1. \(−15x^{3}y\)

3. \(2x−\frac{3}{2}\)

5. \(15x^{2}−6x\)

7. \(3x^{3}+2x^{2}\)

9. \(8a^{2}b−4ab^{2}\)

11. \(−18x^{5}y^{4}+6x^{3}y^{5}\)

13. \(−2x^{3}y^{2}+5x^{2}y\)

15. \(−30x^{9}\)

17. \(−10x^{2}+6x−8\)

19. \(15x^{3}−6x^{2}+9x\)

21. \(3x^{4}−5x^{3}−7x^{2}\)

23. \(2x^{7}−\frac{1}{2}x^{6}+\frac{1}{8}x^{5}−\frac{5}{4}x^{4}\)

25. \(a^{4}b−3a^{3}b^{2}+a^{2}b^{3}\)

27. \(6x^{4}y^{3}−18x^{2}y^{3}+2x^{2}y^{5}\)

29. \(6x^{3}−9x^{2}+15x\)

31. \(−4x^{5}+12x^{4}−8x^{3}+28x^{2}−32x\)

Exercise \(\PageIndex{6}\) Product of a Binomial and a Polynomial

• \((3x−2)(x+4) \)
• \((x+2)(x−3) \)
• \((x−1)(x+1) \)
• \((3x−1)(3x+1) \)
• \((2x−5)(x+3) \)
• \((5x−2)(3x+4) \)
• \((−3x+1)(x−1) \)
• \((x+5)(−x+1) \)
• \((y−\frac{2}{3})(y+\frac{2}{3})\)
• \((\frac{1}{2}x+\frac{1}{3})(\frac{3}{2}x−\frac{2}{3})\)
• \((\frac{3}{4}x+\frac{1}{5})(\frac{1}{4}x+\frac{2}{5})\)
• \((\frac{1}{5}x+\frac{3}{10})(\frac{3}{5}x−\frac{5}{2})\)
• \((y^{2}−2)(y+2)\)
• \((y^{3}−1)(y^{2}+2)\)
• \((a^{2}−b^{2})(a^{2}+b^{2})\)
• \((a^{2}−3b)^{2}\)
• \((x−5)(2x^{2}+3x+4) \)
• \((3x−1)(x^{2}−4x+7) \)
• \((2x−3)(4x^{2}+6x+9) \)
• \((5x+1)(25x^{2}−5x+1) \)
• \((x−\frac{1}{2})(3x^{2}+4x−1) \)
• \((\frac{1}{3}x−\frac{1}{4})(3x^{2}+9x−3) \)
• \((x+3)^{3}\)
• \((x−2)^{3}\)
• \((3x−1)^{3}\)
• \((2x+y)^{3}\)
• \((5x−2)(2x^{3}−4x^{2}+3x−2)\)
• \((x^{2}−2)(x^{3}−2x^{2}+x+1)\)

1. \(3x^{2}+10x−8 \)

3. \(x^{2}−1 \)

5. \(2x^{2}+x−15 \)

7. \(−3x^{2}+4x−1 \)

9. \(y^{2}−\frac{4}{9}\)

11. \(\frac{3}{16}x^{2}+\frac{7}{20}x+\frac{2}{25}\)

13. \(y^{3}+2y^{2}−2y−4\)

15. \(a^{4}−b^{4}\)

17. \(2x^{3}−7x^{2}−11x−20 \)

19. \(8x^{3}−27 \)

21. \(3x^{3}+\frac{5}{2}x^{2}−3x+12\)

23. \(x^{3}+9x^{2}+27x+27\)

25. \(27x^{3}−27x^{2}+9x−1 \)

27. \(10x^{4}−24x^{3}+23x^{2}−16x+4\)

Exercise \(\PageIndex{7}\) Product of Polynomials

• \((x^{2}−x+1)(x^{2}+2x+1)\)
• \((3x^{2}−2x−1)(2x^{2}+3x−4)\)
• \((2x^{2}−3x+5)(x^{2}+5x−1)\)
• \((a+b+c)(a−b−c)\)
• \((a+2b−c)^{2}\)
• \((x+y+z)^{2}\)
• \((x−3)^{4}\)
• \((x+y)^{4}\)
• Find the volume of a rectangular solid with sides measuring \(x, x+2\), and \(x+4\) units.
• Find the volume of a cube where each side measures \(x−5\) units.

1. \(x^{4}+x^{3}+x+1\)

3. \(2x^{4}+7x^{3}−12x^{2}+28x−5\)

5. \(a^{2}+4ab−2ac+4b^{2}−4bc+c^{2}\)

7. \(x^{4}−12x^{3}+54x^{2}−108x+81\)

9. \(x^{3}+6x^{2}+8x\)

Exercise \(\PageIndex{8}\) Special Products

• \((x+2)^{2}\)
• \((x−3)^{2}\)
• \((2x+5)^{2}\)
• \((3x−7)^{2}\)
• \((−x+2)^{2}\)
• \((−9x+1)^{2}\)
• \((a+6)^{2}\)
• \((2a−3b)^{2}\)
• \((\frac{2}{3}x+\frac{3}{4})^{2}\)
• \((\frac{1}{2}x−\frac{3}{5})^{2}\)
• \((x^{2}+2)^{2}\)
• \((x^{2}+y^{2})^{2}\)
• \((x+4)(x−4)\)
• \((2x+1)(2x−1)\)
• \((5x+3)(5x−3)\)
• \((\frac{1}{5}x−\frac{1}{3})(\frac{1}{5}x+\frac{1}{3})\)
• \((\frac{3}{2}x+\frac{2}{5})(\frac{3}{2}x−\frac{2}{5})\)
• \((2x−3y)(2x+3y)\)
• \((4x−y)(4x+y)\)
• \((a^{3}−b^{3})(a^{3}+b^{3})\)

1. \(x^{2}+4x+4 \)

3. \(4x^{2}+20x+25 \)

5. \(x^{2}−4x+4 \)

7. \(a^{2}+12a+36 \)

9. \(\frac{4}{9}x^{2}+x+\frac{9}{16}\)

11. \(x^{4}+4x^{2}+4\)

13. \(x^{2}−16 \)

15. \(25x^{2}−9 \)

17. \(\frac{9}{4}x^{2}−\frac{4}{25}\)

19. \(16x^{2}−y^{2}\)

21. \(V=2x^{2}−16x+32\) cubic inches

Exercise \(\PageIndex{9}\) Multiplying Polynomial Functions

For each problem, calculate \((f⋅g)(x)\), given the functions.

• \(f(x)=8x\) and \(g(x)=3x−5\)
• \(f(x)=x^{2}\) and \(g(x)=−5x+1\)
• \(f(x)=x−7\) and \(g(x)=6x−1\)
• \(f(x)=5x+3\) and \(g(x)=x^{2}+2x−3\)
• \(f(x)=x^{2}+6x−3\) and \(g(x)=2x^{2}−3x+5\)
• \(f(x)=3x^{2}−x+1\) and \(g(x)=−x^{2}+2x−1\)

1. \((f⋅g)(x)=24x^{2}−40x\)

3. \((f⋅g)(x)=6x^{2}−43x+7\)

5. \((f⋅g)(x)=2x^{4}+9x^{3}−19x^{2}+39x−15\)

Exercise \(\PageIndex{10}\) Multiplying Polynomial Functions

Given \(f(x)=2x−3\) and \(g(x)=3x−1\), find the following

• \((f⋅g)(x)\)
• \((g⋅f)(x)\)
• \((f⋅g)(0)\)
• \((f⋅g)(−1)\)
• \((f⋅g)(1)\)
• \((f⋅g)(\frac{1}{2})\)

1. \((f⋅g)(x)=6x^{2}−11x+3\)

3. \((f⋅g)(0)=3\)

5. \((f⋅g)(1)=−2\)

Exercise \(\PageIndex{11}\) Multiplying Polynomial Functions

Given \(f(x)=5x−1\) and \(g(x)=2x^{2}−4x+5\), find the following.

• \((f⋅g)(x) \)
• \((g⋅f)(x) \)
• \((f⋅g)(0) \)
• \((f⋅g)(−1) \)
• \((f⋅g)(1) \)
• \((f⋅f)(x) \)
• \((g⋅g)(x)\)

1. \((f⋅g)(x)=10x^{3}−22x^{2}+29x−5\)

3. \((f⋅g)(0)=−5\)

5. \((f⋅g)(1)=12\)

7. \((f⋅f)(x)=25x^{2}−10x+1\)

Exercise \(\PageIndex{12}\) Discussion Board Topics

• Explain why \((x+y)^{2}\neq x^{2}+y{2}\).
• Explain how to quickly multiply a binomial with its conjugate. Give an example.
• What are the advantages and disadvantages of using the mnemonic device FOIL?

1. Answers may vary

3. Answers may vary

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Multiplication of Polynomials

    Definition and Notation

The product of two natural numbers, 3 and 4 , is defined by

           3 x 4 = 4 + 4 + 4 three terms of 4 Similarly   5a = 5 * a = a + a + a + a + a five terms of a            4ab = ab + ab + ab + ab four terms of ab            ab = a x b = b + b +....+ b a terms of b

The following are some of the laws from multiplication of real numbers

1. The commutative law of multiplication: ab = ba

2. The associative law of multiplication: a(bc) = (ab)c

3. The distributive law of multiplication: a(b + c) = (b + c)a

4. Multiplication of signed numbers:

             +a)(+b) = +ab ; (+a)(-b) = -ab

             (-a)(+b) = -ab ; (-a)(-b) = +ab

When we have 2 * 2 * 2 * 2 . that is, four factors of 2 , the notation 2^4 is used, which reads. “two to the power four," or “two to the fourth power."

Similarly, a * a * a * a * a = a^5 means five factors of a. The a is called the base , and the 5 is called the exponent . When there is no exponent, as in x , we always mean x to the power 1 .

Note the difference between

           (-2^4) = (-2)(-2)(-2)(-2) = + 16

and -2^4 = -(2^4) = - (2 * 2 *2 *2) = -16

Note also 2a^3 = 2(a*a*a)

While (2a)^3 = (2a)(2a)(2a)

= (2*2*2)(a*a*a)

= 2^3a^3 = 8a^3

Remark a , a^2 , a^3 ,.... are not like terms.

EXAMPLES 1. 7a*a*a*a = 7a^4

2. -(-3)(-3)(-3)(-3) = -(-3)^4

3. (x - 1)^3 = (x - 1)(x -1)(x - 1)

4. -2^2*3^3 = -(2*2)*(3*3*3) = -4*27 = -108

5. 2^2 + 2^3 = 2*2 + 2*2*2 = 4 + 8 = 12

6. 2^3 - 2 = 2*2*2 -2 = 8 - 2 = 6

Multiplication of Monomials    

We will discuss the multiplication of monomials, then the multiplication of a monomial and a polynomial, and finally the multiplication of two polynomials.

From the definition of exponents we have

     a^3*a^5 = (a*a*a)(a*a*a*a*a)

  = a*a*a*a*a*a*a*a

  = a^8

  = a^(3+5)

EXAMPLES 1. 2^3 * 2^5 = 2^(3+5) = 2^8

2. a^2 * a^4 = a^(2+4) = a^6

3. -2^4 * 2^3 = -2^(4+3) = -2^7

4. -3x^3*x^2 = -3x^(3+2) = -3x^5

5. x^5*x = x^(5+1) = x^6

6. (a + 1)^2*(a + 1)^3 = (a + 1)^(2+3) = (a + 1)^5

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Remark 2^3*2^7 = 2^(3+7) = 2^10 , and not 4^10

Remark 2^4*3^5 = 2^4*3^5 , to find the product, multiply 2^4 = 16 by 3^5 = 243 ; that is 2^4*3^5 = (16)(243) = 3888

Since the commutative and associative laws for multiplication hold for numbers, specific or general. we have

EXAMPLES 1. (2ab^2)(3a^4bc^2) = (2*3)(a^1*a^4)(b^2*b^1)(c^2)

= 6a^5b^3c^2

2. -3b^2c^3)(8ab^3c) = (-3*8)(b^2*b^3)(c^3*c)(a)

= -24b^5c^4a

3. -3^2xy^2)(-5x^2y^3) = (-9xy^2)(-5x^2y^3)

= (-9)(-5)(x*x^2)(y^2*y^3) = 45x^3y^5  

Let's see how our Polynomial solver simplifies this and similar problems step-by-step. Click on "Solve Similar" button to see more examples.

         (a^2)^3 = (a^2)(a^2)(a^2)

  = (a^2*a^2)(a^2)

  = a^(2+2)*a^2

  = a^(2+2+2)

  = a^(3*2) = a^(2*3)

  = a^6

EXAMPLES 1. (3^2)^4 = 3^(2*4) = 3^8

2. (a^3)^5 = a^(3*5) = a^15

3. (-3^2)^3 = -3^(2*3) = -3^6

4. (-a^3)^2 = a^(3*2) = a^6

Note 2^3*2^4 = 2^(3+4) = 2^7 , While (2^3)^4 = 2^(3*4) = 2^12

   6^4 = (2*3)^4 = (2*3)(2*3)(2*3)(2*3)

  = (2*2*2*2)(3*3*3*3)

  = 2^4*3^4

   Note a and b are factors. If a = 3 , b = x , and m = 5 , (3x)^5 = 3^5x^5

Do not forget to raise the number 3 to the power 5

Applying Theorem 3 repeatedly, we obtain

         (abcd)^m = [(ab)(cd)]^m

  = (ab)^m(cd)^m

  = a^m b^m c^m d^m

     Remark 21^2 = (3*7)^2 = 3^2*7^2 = 9 x 49 = 441

           (5 + 3)^2  = (8)^2 = 64 , but 5^2 + 3^2 = 25 + 9 = 34

If we consider (a + b) as one quantity, then

           (a +b)^5 = (a +b)(a + b)(a + b)(a + b)(a + b)

The method of calculating the product will be explained later

           (a^m)(b^n)^k = [(a^m)(b^n)]^k

= (a^m)^k(b^n)^k

= a^mk b^nk

EXAMPLE Perform the following multiplication: (2x^2yz^3) (-4x^3y^2)

Solution (2x^2yz^3) (-4x^3y^2) = (2)(-4)(x^2*x^3)(y*y^2)(z^3)

= -8x^5y^3z^3

Let's see how our Polynomial solver simplifies this and similar problems, showing explanations for each step. Click on "Solve Similar" button to see more solved examples.

EXAMPLE Perform the following multiplication: (5a^2b)^3

Solution (5a^2b)^3 = (5)^3(a^2)^3(b)^3 = 5^3a^6b^3 = 125a^6b^3

EXAMPLE Perform the following multiplication: -2^2a^3(ab^3)^2

Solution -2^2a^3(ab^3)^2 = -4a^3(a^2b^6) = -4(a^3*a^2)(b^6) = -4a^5b^6

EXAMPLES Perform the following multiplication: (3x^2y)^2(2xy^3)^3

Solution (3x^2y)^2(2xy^3)^3 = (3^2x^4y^2)(2^3x^3y^9) = (3^2*2^3)(x^4*x^3)(y^2*y^9)

= (9*8)x^7y^11 = 72x^7y^11

Remark Perform the outside exponents first

EXAMPLE Perform the following multiplication:

       (-2ab^2)^2(-3a^2b)^3(-bc^2)^4

Solution (-2ab^2)^2(-3a^2b)^3(-bc^2)^4 = (-2)^2a^2b^2*(-3)^3a^6b^3*(-1)^4b^4c^8

= (-2)^2(-3)^3(-1)^4(a^2*a^6)(b^4*b^3*b^4)(c^8)

= (4)(-27)(+1)a^8b^11c^8

= -108a^8b^11c^8

EXAMPLE Perform the indicated operations and simplify:

        (2ab)^4(-a^3b)^2 - (-3a^2)^3(a^2b^3)^2

Solution (2ab)^4(-a^3b)^2 - (-3a^2)^3(a^2b^3)^2 = (16a^4b^4)(a^6b^2) - (-27a^6)(a^4b^6)

= 16a^10b^6 + 27a^10b^6

= 43a^10b^6

Let's see how our Polynomial calculator explains all simplification steps for this and similar problems. Click on "Solve Similar" button to see more examples.

Note : To evaluate expressions involving exponents. first replace each letter by its indicated specific value. Use grouping symbols where necessary so as not to confuse operation signs with number signs.

Evaluate -a^2b^3 , give that a = -3 and b = 2

Solution -a^2b^3 = -(-3)^2(2)^3 = -(9)(8) = -72

EXAMPLE Evaluate the expression b^2 - a^2(c^3 - b^3) , give that a = -2 , b = 3 , and c = -1

Solution b^2 -a^2(c^3 - b^3) = (3)^2 - (-2)^2[((-1)^3 - (3)^3

= 9 - (+4)[(-1) - (27)

= 9 - 4(-1 -27)

= 9 - 4(-28)

= 9 + 112 = 121

Multiplication of a Polynomial by a Monomial     

The extended distributive law of multiplication

is used to multiply a monomial by a polynomial

EXAMPLE Multiply 3x^2 + x - 2 by x

Solution x(3x^2 + x - 2) = x(3x^2) + x(x) + x(-2)

= 3x^2 + x^2 - 2x

EXAMPLE Multiply x^2 - x + 4 by -2x^2

Solution (-2x^2)(x^2 - x + 4) = (-2x^2)(x^2) + (-2x^2)(-x) + (-2x^2)(4)

= -2x^4 + 2x^3 - 8x^2

EXAMPLE Multiply a^2b - 2b^2c + 5c^2a by 3a^2b

Solution 3a^2b(a^2b - 2b^2c + 5c^2a)

= 3a^2b(a^2b) + 3a^2b(-2b^2c) + 3a^2b(5c^2a)

= 3a^4b^2 - 6a^2b^3c + 15a^3bc^2

EXAMPLE Multiply (3x - 2)/4 - (2x - 1)/6 by 12

Solution 12/1 [(3x - 2)/4 - (2x - 1)/6 = 12/1 [(3x - 2)/4 - 12/1[(2x - 1)/6

= 3(3x - 2) - 2(2x - 1)

= 9x - 6 - 4x + 2

Multiplication of two polynomials is the same as multiplication of a monomial and a polynomial where the first polynomial is considered as one quantity.

To multiply (x + 2) by (x - 3) , consider (x + 2) as one quantity and apply the distributive law:

Then reapply the distributive law

= x^2 + 2x - 3x - 6

= x^2 - x - 6

Notice that each term of the second polynomial has been multiplied by each term of the first polynomial.

The same result can be obtained by arranging the polynomials in two rows and multiplying the upper polynomial by each term of the lower polynomial. Arrange like terms of the product in the same column so that addition is easier.+

EXAMPLE Multiply (3x - 4)^2

Solution (3x - 4)^2 = (3x - 4)(3x - 4)

Notes 1. (a + b)^2 = a^2 + 2ab + b^2

2. (a - b)^2 = a^2 - 2ab + b^2

3. (a + b)(a - b) = a^2 - b^2

EXAMPLE Multiply (x^2 - 2x +1) by (2x - 3)

Solution   

Therefore (x^2 - 2x + 1)(2x - 3) = 2x^3 - 7x^2 + 8x - 3

EXAMPLE Perform the indicated operations and simplify

    (2x - 3)(x + 4) - (x + 2)(x - 6)

Solution (2x - 3)(x + 4) - (x + 2)(x - 6)

= (2x^2 -+5x -12) - (x^2 - 4x - 12)

= 2x^2 + 5x - 12 - x^2 + 4x + 12

Grouping Symbols

Grouping symbols, such as parentheses ( ), braces { }, and brackets [ ], are used to designate, in a simple manner, more than one operation.

When we write the binomial 3a + 5b as (3a + 5b) , we are considering the sum of 3a and 5b as one quantity. The expression a - (b + c) means the sum of b and c is to be subtracted from a .

The statement, three times x minus four times the sum of y and z , can be written in algebraic notation as

3x - 4 (y + z)

Removal of the grouping symbols means performing the operations that these symbols indicate. Remove the symbols one at a time, starting with the innermost, following the proper order of operations to be performed.

EXAMPLE Remove the grouping symbols and combine like terms

         2x - (5x - 2y) + (x - 6y)

Solution 2x - (5x - 2y) + (x - 6y) = 2x - 5x + 2y + x -6y

  = (2x - 5x + x) + (2y - 6y)

  = -2x - 4y

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         7a + 2[2b - 3(3a - 5b)

Solution 7a + 2[2b - 3(3a - 5b) = 7a + 2[2b - 9a +15b

  = 7a + 4b -18a +30b

  = 34b -11a

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         6a - {2b + [3 - (a + b) + (5a - 2)}

Solution 6a - {2b + [3 - (a + b) + (5a - 2)}

= 6a - {2b + [3 - a - b + 5a -2]}

= 6a - {2b + 3 - a -b +5a -2}

= 6a - 2b -3 +a +b -5a +2

= (6a + a - 5a) + (-2b + b) + (-3 +2)

= 2a - b -1

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It is sometimes necessary to group some of the terms of an expression. This can be accomplished by use of a set of parentheses.

When the grouping symbol is preceded by a plus sign. we keep the signs of the terms the same when it is preceded by a minus sign, we use the additive inverses (negatives) of the terms.

EXAMPLE Group the last three terms of the polynomial 3a -  5b + c -  2 with a grouping symbol in two ways, one preceded by a plus sign, the second preceded by a minus sign.

Math Topics

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• Add Fractions
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Multiplying Polynomials

A polynomial looks like this:

To multiply two polynomials:

• multiply each term in one polynomial by each term in the other polynomial
• add those answers together, and simplify if needed

Let us look at the simplest cases first.

1 term × 1 term   (monomial times monomial)

To multiply one term by another term, first multiply the constants , then multiply each variable together and combine the result, like this (press play):

(Note: I used "·" to mean multiply. In Algebra we don't like to use "×" because it looks too much like the letter "x")

For more about multiplying terms, read Multiply and Divide Variables with Exponents

1 term × 2 terms   (monomial times binomial)

Multiply the single term by each of the two terms, like this:

2 term × 1 terms   (binomial times monomial)

Multiply each of the two terms by the single term, like this:

(I did that one a bit faster by multiplying in my head before writing it down)

2 terms × 2 terms (binomial times binomial)

That is 4 different multiplications ... Why?

It is the same when we multiply binomials!

Instead of Alice and Betty, let's just use a and b , and Charles and David can be c and d :

We can multiply them in any order so long as each of the first two terms gets multiplied by each of the second two terms .

But there is a handy way to help us remember to multiply each term called " FOIL ".

It stands for " F irsts, O uters, I nners, L asts":

So you multiply the "Firsts" (the first terms of both polynomials), then the "Outers", etc.

Let us try this on a more complicated example:

2 terms × 3 terms (binomial times trinomial)

"FOIL" won't work here, because there are more terms now. But just remember:

Multiply each term in the first polynomial by each term in the second polynomial

And always remember to add Like Terms :

Example: (x + 2y)(3x − 4y + 5)

(x + 2y)(3x − 4y + 5)

= 3x 2 − 4xy + 5x + 6xy − 8y 2 + 10y

= 3x 2 + 2xy + 5x − 8y 2 + 10y

Note: −4xy and 6xy are added because they are Like Terms.

Also note: 6yx means the same thing as 6xy

Long Multiplication

You may also like to read about Polynomial Long Multiplication

• Math Article

Multiplying Polynomials

Multiplying polynomials is a basic concept in algebra. Multiplication of two polynomials will include the product of coefficients to coefficients and variables to variables. We can easily multiply polynomials using rules and following some simple steps. Let us learn more about multiplying polynomials with examples in this article.

Rules for Multiplying Polynomials

Multiplying polynomials require only three steps.

• First, multiply each term in one polynomial by each term in the other polynomial using the distributive law.
• Add the powers of the same variables using the exponent rule.
• Then, simplify the resulting polynomial by adding or subtracting the like terms.

It should be noted that the resulting degree after multiplying two polynomials will be always more than the degree of the individual polynomials.

Multiplying Polynomials Using Exponent Law

If the variable is the same but has different exponents of the given polynomials, then we need to use the exponent law.

Example: Multiply 2x 2  × 3x

Here, the coefficients and variables are multiplied separately.

= (2 × 3) × (x 2  × x)

= 6 × x 2+1

Multiplying Polynomials having different variables

Follow the below-given steps for multiplying polynomials:

• Step 1: Place the two polynomials in a line.

For example, for two polynomials, (6x−3y) and (2x+5y), write as: (6x−3y)×(2x+5y)

• Step 2: Use distributive law and separate the first polynomial.
• Step 3: Multiply the monomials from the first polynomial with each term of the second polynomial.

⇒ [6x × (2x+5y)] − [3y × (2x+5y)] = (12x 2 +30xy) − (6yx+15y 2 )

• Step 4: Simplify the resultant polynomial, if possible.

⇒ (12x 2 +30xy) − (6yx+15y 2 ) = 12x 2 +24xy−15y 2

Degree – Multiplying Polynomials

For two polynomials equations, P and Q, the degree after multiplication will always be higher than the degree of P or Q. The degree of the resulting polynomial will be the summation of the degree of P and Q.

Multiplying Polynomials by Polynomials

It is known that there are different types of polynomial based on their degree like monomial, binomial, trinomial, etc. The steps to multiply polynomials are the same for all types.

Multiplying Monomial by Monomial

A monomial is a single term polynomial. If two or more monomials are multiplied together, then the resulting product will be a monomial.

Examples are:

• 5x × 7x = 5 × x × 7 × x = 35x 2
• 2x × 3y × 4z = 2 × x × 3 × y × 4 × z = (2 × 3 × 4) × (x × y × z) = 24xyz

Multiplying Binomial by a Binomial

A binomial is a two-term polynomial. When a binomial is multiplied by a binomial, the distributive law of multiplication is followed.

We know that Binomial has 2 terms. Multiplying two binomials give the result having a maximum of 4 terms (only in case when we don’t have like terms). In the case of like terms, the total number of terms is reduced.

Like Terms:

According to the commutative law of multiplication, terms like ‘ab’ and ‘ba’ give the same result. Thus they can be written in both forms.

For example, 5×6 = 6×5 = 30

Now, Consider two binomials given as (a+b) and (m+n).

Multiplying them we have,

(a+b)×(m+n)

⇒ a×(m+n)+b×(m+n) (Distributive law of multiplication)

⇒ (am+an)+(bm+bn) (Distributive law of multiplication)

Solved Examples

Example 1: Find the result of multiplication of two polynomials (6x +3y) and (2x+ 5y).

Solution- (6x−3y)×(2x+5y)

⇒6x×(2x+5y)−3y×(2x+5y) (Distributive law of multiplication)

⇒(12x 2 +30xy)−(6yx+15y 2 ) (Distributive law of multiplication)

⇒12x 2 +30xy−6xy−15y 2 (as xy = yx)

Thus, (6x+3y)×(2x+5y)=12x 2 +24xy−15y 2

Let us take up an example. Say, you are required to multiply a binomial (5y + 3z) with another binomial (7y − 15z). Let us see how it is done.

(5y + 3z) × (7y − 15z) = 5y × (7y − 15z) + 3z × (7y − 15z) (Distributive law of multiplication)

= (5y × 7y) − (5y × 15z) + (3z × 7y) − (3z × 15z) (Distributive law of multiplication)

= 35y 2 − 75yz + 21zy − 45z 2

= 35y 2 − 75yz + 21yz − 45z 2

As, (yz = zy)

(5y + 3z) × (7y − 15z) = 35y 2 −54yz − 45z 2

Multiplying Binomial with a Trinomial

A trinomial is a three-term polynomial. When multiplying polynomials, that is, a binomial by a trinomial, we follow the distributive law of multiplication. Thus, 2 × 3 = 6 terms are expected to be in the product. Let us take up an example.

(a 2 − 2a) × (a + 2b − 3c)

= a 2 × (a + 2b − 3c) − 2a × (a + 2b − 3c) (Distributive law of multiplication)

= (a 2 × a) + (a 2 × 2b) + (a 2 × −3c) − (2a × a) − (2a × 2b) − (2a × −3c) (Distributive law of multiplication)

= a 3 + 2a 2 b − 3a 2 c − 2a 2 − 4ab + 6ac

Now, by rearranging the terms,

(a 2 − 2a) × (a + 2b − 3c) = a 3 − 2a 2 + 2a 2 b − 3a 2 c− 4ab + 6ac

Important Facts

When multiplying polynomials, the following pointers should be kept in mind:

• Distributive Law of multiplication is used twice when 2 polynomials are multiplied.
• Look for the like terms and combine them. This may reduce the expected number of terms in the product.
• Preferably, write the terms in the decreasing order of their exponent.
• Be very careful with the signs when you open the brackets.

Related Articles

Practice questions.

• Multiply 2x by 3y.
• Multiply (3x – a) (4x – y)
• Find the product of (x + 2y)(3x − 4y + 5)
• Multiply (x – 3) (2x – 9)
• What is the product of 3x 2 and 4x 2 – 5x + 7?

Frequently Asked Questions – FAQs

How to multiply polynomials, how to multiply binomials, how to multiply large polynomials.

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6.3 Multiply Polynomials

Learning objectives.

By the end of this section, you will be able to:

• Multiply a polynomial by a monomial
• Multiply a binomial by a binomial
• Multiply a trinomial by a binomial

Be Prepared 6.3

Before you get started, take this readiness quiz.

• Distribute: 2 ( x + 3 ) . 2 ( x + 3 ) . If you missed this problem, review Example 1.132 .
• Combine like terms: x 2 + 9 x + 7 x + 63 . x 2 + 9 x + 7 x + 63 . If you missed this problem, review Example 1.24 .

Multiply a Polynomial by a Monomial

We have used the Distributive Property to simplify expressions like 2 ( x − 3 ) 2 ( x − 3 ) . You multiplied both terms in the parentheses, x and 3 x and 3 , by 2, to get 2 x − 6 2 x − 6 . With this chapter’s new vocabulary, you can say you were multiplying a binomial, x − 3 x − 3 , by a monomial, 2.

Multiplying a binomial by a monomial is nothing new for you! Here’s an example:

Example 6.28

Multiply: 4 ( x + 3 ) . 4 ( x + 3 ) .

Try It 6.55

Multiply: 5 ( x + 7 ) . 5 ( x + 7 ) .

Try It 6.56

Multiply: 3 ( y + 13 ) . 3 ( y + 13 ) .

Example 6.29

Multiply: y ( y − 2 ) . y ( y − 2 ) .

Try It 6.57

Multiply: x ( x − 7 ) . x ( x − 7 ) .

Try It 6.58

Multiply: d ( d − 11 ) . d ( d − 11 ) .

Example 6.30

Multiply: 7 x ( 2 x + y ) . 7 x ( 2 x + y ) .

Try It 6.59

Multiply: 5 x ( x + 4 y ) . 5 x ( x + 4 y ) .

Try It 6.60

Multiply: 2 p ( 6 p + r ) . 2 p ( 6 p + r ) .

Example 6.31

Multiply: −2 y ( 4 y 2 + 3 y − 5 ) . −2 y ( 4 y 2 + 3 y − 5 ) .

Try It 6.61

Multiply: −3 y ( 5 y 2 + 8 y − 7 ) . −3 y ( 5 y 2 + 8 y − 7 ) .

Try It 6.62

Multiply: 4 x 2 ( 2 x 2 − 3 x + 5 ) . 4 x 2 ( 2 x 2 − 3 x + 5 ) .

Example 6.32

Multiply: 2 x 3 ( x 2 − 8 x + 1 ) . 2 x 3 ( x 2 − 8 x + 1 ) .

Try It 6.63

Multiply: 4 x ( 3 x 2 − 5 x + 3 ) . 4 x ( 3 x 2 − 5 x + 3 ) .

Try It 6.64

Multiply: −6 a 3 ( 3 a 2 − 2 a + 6 ) . −6 a 3 ( 3 a 2 − 2 a + 6 ) .

Example 6.33

Multiply: ( x + 3 ) p . ( x + 3 ) p .

Try It 6.65

Multiply: ( x + 8 ) p . ( x + 8 ) p .

Try It 6.66

Multiply: ( a + 4 ) p . ( a + 4 ) p .

Multiply a Binomial by a Binomial

Just like there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a binomial times a binomial. We will start by using the Distributive Property.

Multiply a Binomial by a Binomial Using the Distributive Property

Look at Example 6.33 , where we multiplied a binomial by a monomial .

Notice that before combining like terms, you had four terms. You multiplied the two terms of the first binomial by the two terms of the second binomial—four multiplications.

Example 6.34

Multiply: ( y + 5 ) ( y + 8 ) . ( y + 5 ) ( y + 8 ) .

Try It 6.67

Multiply: ( x + 8 ) ( x + 9 ) . ( x + 8 ) ( x + 9 ) .

Try It 6.68

Multiply: ( 5 x + 9 ) ( 4 x + 3 ) . ( 5 x + 9 ) ( 4 x + 3 ) .

Example 6.35

Multiply: ( 2 y + 5 ) ( 3 y + 4 ) . ( 2 y + 5 ) ( 3 y + 4 ) .

Try It 6.69

Multiply: ( 3 b + 5 ) ( 4 b + 6 ) . ( 3 b + 5 ) ( 4 b + 6 ) .

Try It 6.70

Multiply: ( a + 10 ) ( a + 7 ) . ( a + 10 ) ( a + 7 ) .

Example 6.36

Multiply: ( 4 y + 3 ) ( 2 y − 5 ) . ( 4 y + 3 ) ( 2 y − 5 ) .

Try It 6.71

Multiply: ( 5 y + 2 ) ( 6 y − 3 ) . ( 5 y + 2 ) ( 6 y − 3 ) .

Try It 6.72

Multiply: ( 3 c + 4 ) ( 5 c − 2 ) . ( 3 c + 4 ) ( 5 c − 2 ) .

Example 6.37

Multiply: ( x - 2 ) ( x − y ) . ( x - 2 ) ( x − y ) .

Try It 6.73

Multiply: ( a + 7 ) ( a − b ) . ( a + 7 ) ( a − b ) .

Try It 6.74

Multiply: ( x + 5 ) ( x − y ) . ( x + 5 ) ( x − y ) .

Multiply a Binomial by a Binomial Using the FOIL Method

Remember that when you multiply a binomial by a binomial you get four terms. Sometimes you can combine like terms to get a trinomial , but sometimes, like in Example 6.37 , there are no like terms to combine.

Let’s look at the last example again and pay particular attention to how we got the four terms.

Where did the first term, x 2 x 2 , come from?

We abbreviate “First, Outer, Inner, Last” as FOIL. The letters stand for ‘ F irst, O uter, I nner, L ast’. The word FOIL is easy to remember and ensures we find all four products.

Let’s look at ( x + 3 ) ( x + 7 ) ( x + 3 ) ( x + 7 ) .

Notice how the terms in third line fit the FOIL pattern.

Now we will do an example where we use the FOIL pattern to multiply two binomials.

Example 6.38

How to multiply a binomial by a binomial using the foil method.

Multiply using the FOIL method: ( x + 5 ) ( x + 9 ) . ( x + 5 ) ( x + 9 ) .

Try It 6.75

Multiply using the FOIL method: ( x + 6 ) ( x + 8 ) . ( x + 6 ) ( x + 8 ) .

Try It 6.76

Multiply using the FOIL method: ( y + 17 ) ( y + 3 ) . ( y + 17 ) ( y + 3 ) .

We summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!

Multiply two binomials using the FOIL method

When you multiply by the FOIL method, drawing the lines will help your brain focus on the pattern and make it easier to apply.

Example 6.39

Multiply: ( y − 7 ) ( y + 4 ) . ( y − 7 ) ( y + 4 ) .

Try It 6.77

Multiply: ( x − 7 ) ( x + 5 ) . ( x − 7 ) ( x + 5 ) .

Try It 6.78

Multiply: ( b − 3 ) ( b + 6 ) . ( b − 3 ) ( b + 6 ) .

Example 6.40

Multiply: ( 4 x + 3 ) ( 2 x − 5 ) . ( 4 x + 3 ) ( 2 x − 5 ) .

Try It 6.79

Multiply: ( 3 x + 7 ) ( 5 x − 2 ) . ( 3 x + 7 ) ( 5 x − 2 ) .

Try It 6.80

Multiply: ( 4 y + 5 ) ( 4 y − 10 ) . ( 4 y + 5 ) ( 4 y − 10 ) .

The final products in the last four examples were trinomials because we could combine the two middle terms. This is not always the case.

Example 6.41

Multiply: ( 3 x − y ) ( 2 x − 5 ) . ( 3 x − y ) ( 2 x − 5 ) .

Try It 6.81

Multiply: ( 10 c − d ) ( c − 6 ) . ( 10 c − d ) ( c − 6 ) .

Try It 6.82

Multiply: ( 7 x − y ) ( 2 x − 5 ) . ( 7 x − y ) ( 2 x − 5 ) .

Be careful of the exponents in the next example.

Example 6.42

Multiply: ( n 2 + 4 ) ( n − 1 ) . ( n 2 + 4 ) ( n − 1 ) .

Try It 6.83

Multiply: ( x 2 + 6 ) ( x − 8 ) . ( x 2 + 6 ) ( x − 8 ) .

Try It 6.84

Multiply: ( y 2 + 7 ) ( y − 9 ) . ( y 2 + 7 ) ( y − 9 ) .

Example 6.43

Multiply: ( 3 p q + 5 ) ( 6 p q − 11 ) . ( 3 p q + 5 ) ( 6 p q − 11 ) .

Try It 6.85

Multiply: ( 2 a b + 5 ) ( 4 a b − 4 ) . ( 2 a b + 5 ) ( 4 a b − 4 ) .

Try It 6.86

Multiply: ( 2 x y + 3 ) ( 4 x y − 5 ) . ( 2 x y + 3 ) ( 4 x y − 5 ) .

Multiply a Binomial by a Binomial Using the Vertical Method

The FOIL method is usually the quickest method for multiplying two binomials, but it only works for binomials. You can use the Distributive Property to find the product of any two polynomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.

Now we’ll apply this same method to multiply two binomials.

Example 6.44

Multiply using the Vertical Method: ( 3 y − 1 ) ( 2 y − 6 ) . ( 3 y − 1 ) ( 2 y − 6 ) .

It does not matter which binomial goes on the top.

Multiply 3 y − 1 by −6 . Multiply 3 y − 1 by 2y. Add like terms. 3 y − 1 × 2 y − 6 ________ −18 y + 6 6 y 2 − 2 y  _____________ 6 y 2 − 20 y + 6 partial product partial product product Multiply 3 y − 1 by −6 . Multiply 3 y − 1 by 2y. Add like terms. 3 y − 1 × 2 y − 6 ________ −18 y + 6 6 y 2 − 2 y  _____________ 6 y 2 − 20 y + 6 partial product partial product product

Notice the partial products are the same as the terms in the FOIL method.

Try It 6.87

Multiply using the Vertical Method: ( 5 m − 7 ) ( 3 m − 6 ) . ( 5 m − 7 ) ( 3 m − 6 ) .

Try It 6.88

Multiply using the Vertical Method: ( 6 b − 5 ) ( 7 b − 3 ) . ( 6 b − 5 ) ( 7 b − 3 ) .

We have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The methods are listed here all together, to help you remember them.

Multiplying Two Binomials

To multiply binomials, use the:

• Distributive Property
• FOIL Method
• Vertical Method

Remember, FOIL only works when multiplying two binomials.

Multiply a Trinomial by a Binomial

We have multiplied monomials by monomials, monomials by polynomials, and binomials by binomials. Now we’re ready to multiply a trinomial by a binomial . Remember, FOIL will not work in this case, but we can use either the Distributive Property or the Vertical Method. We first look at an example using the Distributive Property.

Example 6.45

Multiply using the Distributive Property: ( b + 3 ) ( 2 b 2 − 5 b + 8 ) . ( b + 3 ) ( 2 b 2 − 5 b + 8 ) .

Try It 6.89

Multiply using the Distributive Property: ( y − 3 ) ( y 2 − 5 y + 2 ) . ( y − 3 ) ( y 2 − 5 y + 2 ) .

Try It 6.90

Multiply using the Distributive Property: ( x + 4 ) ( 2 x 2 − 3 x + 5 ) . ( x + 4 ) ( 2 x 2 − 3 x + 5 ) .

Now let’s do this same multiplication using the Vertical Method.

Example 6.46

Multiply using the Vertical Method: ( b + 3 ) ( 2 b 2 − 5 b + 8 ) . ( b + 3 ) ( 2 b 2 − 5 b + 8 ) .

It is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products this way.

Try It 6.91

Multiply using the Vertical Method: ( y − 3 ) ( y 2 − 5 y + 2 ) . ( y − 3 ) ( y 2 − 5 y + 2 ) .

Try It 6.92

Multiply using the Vertical Method: ( x + 4 ) ( 2 x 2 − 3 x + 5 ) . ( x + 4 ) ( 2 x 2 − 3 x + 5 ) .

We have now seen two methods you can use to multiply a trinomial by a binomial. After you practice each method, you’ll probably find you prefer one way over the other. We list both methods are listed here, for easy reference.

Multiplying a Trinomial by a Binomial

To multiply a trinomial by a binomial, use the:

Access these online resources for additional instruction and practice with multiplying polynomials:

• Multiplying Exponents 1
• Multiplying Exponents 2
• Multiplying Exponents 3

Section 6.3 Exercises

Practice makes perfect.

In the following exercises, multiply.

4 ( w + 10 ) 4 ( w + 10 )

6 ( b + 8 ) 6 ( b + 8 )

−3 ( a + 7 ) −3 ( a + 7 )

−5 ( p + 9 ) −5 ( p + 9 )

2 ( x − 7 ) 2 ( x − 7 )

7 ( y − 4 ) 7 ( y − 4 )

−3 ( k − 4 ) −3 ( k − 4 )

−8 ( j − 5 ) −8 ( j − 5 )

q ( q + 5 ) q ( q + 5 )

k ( k + 7 ) k ( k + 7 )

− b ( b + 9 ) − b ( b + 9 )

− y ( y + 3 ) − y ( y + 3 )

− x ( x − 10 ) − x ( x − 10 )

− p ( p − 15 ) − p ( p − 15 )

6 r ( 4 r + s ) 6 r ( 4 r + s )

5 c ( 9 c + d ) 5 c ( 9 c + d )

12 x ( x − 10 ) 12 x ( x − 10 )

9 m ( m − 11 ) 9 m ( m − 11 )

−9 a ( 3 a + 5 ) −9 a ( 3 a + 5 )

−4 p ( 2 p + 7 ) −4 p ( 2 p + 7 )

3 ( p 2 + 10 p + 25 ) 3 ( p 2 + 10 p + 25 )

6 ( y 2 + 8 y + 16 ) 6 ( y 2 + 8 y + 16 )

−8 x ( x 2 + 2 x − 15 ) −8 x ( x 2 + 2 x − 15 )

−5 t ( t 2 + 3 t − 18 ) −5 t ( t 2 + 3 t − 18 )

5 q 3 ( q 3 − 2 q + 6 ) 5 q 3 ( q 3 − 2 q + 6 )

4 x 3 ( x 4 − 3 x + 7 ) 4 x 3 ( x 4 − 3 x + 7 )

−8 y ( y 2 + 2 y − 15 ) −8 y ( y 2 + 2 y − 15 )

−5 m ( m 2 + 3 m − 18 ) −5 m ( m 2 + 3 m − 18 )

5 q 3 ( q 2 − 2 q + 6 ) 5 q 3 ( q 2 − 2 q + 6 )

9 r 3 ( r 2 − 3 r + 5 ) 9 r 3 ( r 2 − 3 r + 5 )

−4 z 2 ( 3 z 2 + 12 z − 1 ) −4 z 2 ( 3 z 2 + 12 z − 1 )

−3 x 2 ( 7 x 2 + 10 x − 1 ) −3 x 2 ( 7 x 2 + 10 x − 1 )

( 2 m − 9 ) m ( 2 m − 9 ) m

( 8 j − 1 ) j ( 8 j − 1 ) j

( w − 6 ) · 8 ( w − 6 ) · 8

( k − 4 ) · 5 ( k − 4 ) · 5

4 ( x + 10 ) 4 ( x + 10 )

6 ( y + 8 ) 6 ( y + 8 )

15 ( r − 24 ) 15 ( r − 24 )

12 ( v − 30 ) 12 ( v − 30 )

−3 ( m + 11 ) −3 ( m + 11 )

−4 ( p + 15 ) −4 ( p + 15 )

−8 ( z − 5 ) −8 ( z − 5 )

−3 ( x − 9 ) −3 ( x − 9 )

u ( u + 5 ) u ( u + 5 )

q ( q + 7 ) q ( q + 7 )

n ( n 2 − 3 n ) n ( n 2 − 3 n )

s ( s 2 − 6 s ) s ( s 2 − 6 s )

6 x ( 4 x + y ) 6 x ( 4 x + y )

5 a ( 9 a + b ) 5 a ( 9 a + b )

5 p ( 11 p − 5 q ) 5 p ( 11 p − 5 q )

12 u ( 3 u − 4 v ) 12 u ( 3 u − 4 v )

3 ( v 2 + 10 v + 25 ) 3 ( v 2 + 10 v + 25 )

6 ( x 2 + 8 x + 16 ) 6 ( x 2 + 8 x + 16 )

2 n ( 4 n 2 − 4 n + 1 ) 2 n ( 4 n 2 − 4 n + 1 )

3 r ( 2 r 2 − 6 r + 2 ) 3 r ( 2 r 2 − 6 r + 2 )

( 2 y − 9 ) y ( 2 y − 9 ) y

( 8 b − 1 ) b ( 8 b − 1 ) b

In the following exercises, multiply the following binomials using: ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical Method.

( w + 5 ) ( w + 7 ) ( w + 5 ) ( w + 7 )

( y + 9 ) ( y + 3 ) ( y + 9 ) ( y + 3 )

( p + 11 ) ( p − 4 ) ( p + 11 ) ( p − 4 )

( q + 4 ) ( q − 8 ) ( q + 4 ) ( q − 8 )

In the following exercises, multiply the binomials. Use any method.

( x + 8 ) ( x + 3 ) ( x + 8 ) ( x + 3 )

( y + 7 ) ( y + 4 ) ( y + 7 ) ( y + 4 )

( y − 6 ) ( y − 2 ) ( y − 6 ) ( y − 2 )

( x − 7 ) ( x − 2 ) ( x − 7 ) ( x − 2 )

( w − 4 ) ( w + 7 ) ( w − 4 ) ( w + 7 )

( q − 5 ) ( q + 8 ) ( q − 5 ) ( q + 8 )

( p + 12 ) ( p − 5 ) ( p + 12 ) ( p − 5 )

( m + 11 ) ( m − 4 ) ( m + 11 ) ( m − 4 )

( 6 p + 5 ) ( p + 1 ) ( 6 p + 5 ) ( p + 1 )

( 7 m + 1 ) ( m + 3 ) ( 7 m + 1 ) ( m + 3 )

( 2 t − 9 ) ( 10 t + 1 ) ( 2 t − 9 ) ( 10 t + 1 )

( 3 r − 8 ) ( 11 r + 1 ) ( 3 r − 8 ) ( 11 r + 1 )

( 5 x − y ) ( 3 x − 6 ) ( 5 x − y ) ( 3 x − 6 )

( 10 a − b ) ( 3 a − 4 ) ( 10 a − b ) ( 3 a − 4 )

( a + b ) ( 2 a + 3 b ) ( a + b ) ( 2 a + 3 b )

( r + s ) ( 3 r + 2 s ) ( r + s ) ( 3 r + 2 s )

( 4 z − y ) ( z − 6 ) ( 4 z − y ) ( z − 6 )

( 5 x − y ) ( x − 4 ) ( 5 x − y ) ( x − 4 )

( x 2 + 3 ) ( x + 2 ) ( x 2 + 3 ) ( x + 2 )

( y 2 − 4 ) ( y + 3 ) ( y 2 − 4 ) ( y + 3 )

( x 2 + 8 ) ( x 2 − 5 ) ( x 2 + 8 ) ( x 2 − 5 )

( y 2 − 7 ) ( y 2 − 4 ) ( y 2 − 7 ) ( y 2 − 4 )

( 5 a b − 1 ) ( 2 a b + 3 ) ( 5 a b − 1 ) ( 2 a b + 3 )

( 2 x y + 3 ) ( 3 x y + 2 ) ( 2 x y + 3 ) ( 3 x y + 2 )

( 6 p q − 3 ) ( 4 p q − 5 ) ( 6 p q − 3 ) ( 4 p q − 5 )

( 3 r s − 7 ) ( 3 r s − 4 ) ( 3 r s − 7 ) ( 3 r s − 4 )

In the following exercises, multiply using ⓐ the Distributive Property ⓑ the Vertical Method.

( x + 5 ) ( x 2 + 4 x + 3 ) ( x + 5 ) ( x 2 + 4 x + 3 )

( u + 4 ) ( u 2 + 3 u + 2 ) ( u + 4 ) ( u 2 + 3 u + 2 )

( y + 8 ) ( 4 y 2 + y − 7 ) ( y + 8 ) ( 4 y 2 + y − 7 )

( a + 10 ) ( 3 a 2 + a − 5 ) ( a + 10 ) ( 3 a 2 + a − 5 )

In the following exercises, multiply. Use either method.

( w − 7 ) ( w 2 − 9 w + 10 ) ( w − 7 ) ( w 2 − 9 w + 10 )

( p − 4 ) ( p 2 − 6 p + 9 ) ( p − 4 ) ( p 2 − 6 p + 9 )

( 3 q + 1 ) ( q 2 − 4 q − 5 ) ( 3 q + 1 ) ( q 2 − 4 q − 5 )

( 6 r + 1 ) ( r 2 − 7 r − 9 ) ( 6 r + 1 ) ( r 2 − 7 r − 9 )

Mixed Practice

( 10 y − 6 ) + ( 4 y − 7 ) ( 10 y − 6 ) + ( 4 y − 7 )

( 15 p − 4 ) + ( 3 p − 5 ) ( 15 p − 4 ) + ( 3 p − 5 )

( x 2 − 4 x − 34 ) − ( x 2 + 7 x − 6 ) ( x 2 − 4 x − 34 ) − ( x 2 + 7 x − 6 )

( j 2 − 8 j − 27 ) − ( j 2 + 2 j − 12 ) ( j 2 − 8 j − 27 ) − ( j 2 + 2 j − 12 )

5 q ( 3 q 2 − 6 q + 11 ) 5 q ( 3 q 2 − 6 q + 11 )

8 t ( 2 t 2 − 5 t + 6 ) 8 t ( 2 t 2 − 5 t + 6 )

( s − 7 ) ( s + 9 ) ( s − 7 ) ( s + 9 )

( x − 5 ) ( x + 13 ) ( x − 5 ) ( x + 13 )

( y 2 − 2 y ) ( y + 1 ) ( y 2 − 2 y ) ( y + 1 )

( a 2 − 3 a ) ( 4 a + 5 ) ( a 2 − 3 a ) ( 4 a + 5 )

( 3 n − 4 ) ( n 2 + n − 7 ) ( 3 n − 4 ) ( n 2 + n − 7 )

( 6 k − 1 ) ( k 2 + 2 k − 4 ) ( 6 k − 1 ) ( k 2 + 2 k − 4 )

( 7 p + 10 ) ( 7 p − 10 ) ( 7 p + 10 ) ( 7 p − 10 )

( 3 y + 8 ) ( 3 y − 8 ) ( 3 y + 8 ) ( 3 y − 8 )

( 4 m 2 − 3 m − 7 ) m 2 ( 4 m 2 − 3 m − 7 ) m 2

( 15 c 2 − 4 c + 5 ) c 4 ( 15 c 2 − 4 c + 5 ) c 4

( 5 a + 7 b ) ( 5 a + 7 b ) ( 5 a + 7 b ) ( 5 a + 7 b )

( 3 x − 11 y ) ( 3 x − 11 y ) ( 3 x − 11 y ) ( 3 x − 11 y )

( 4 y + 12 z ) ( 4 y − 12 z ) ( 4 y + 12 z ) ( 4 y − 12 z )

Everyday Math

Mental math You can use binomial multiplication to multiply numbers without a calculator. Say you need to multiply 13 times 15. Think of 13 as 10 + 3 10 + 3 and 15 as 10 + 5 10 + 5 .

• ⓐ Multiply ( 10 + 3 ) ( 10 + 5 ) ( 10 + 3 ) ( 10 + 5 ) by the FOIL method.
• ⓑ Multiply 13 · 15 13 · 15 without using a calculator.
• ⓒ Which way is easier for you? Why?

Mental math You can use binomial multiplication to multiply numbers without a calculator. Say you need to multiply 18 times 17. Think of 18 as 20 − 2 20 − 2 and 17 as 20 − 3 20 − 3 .

• ⓐ Multiply ( 20 − 2 ) ( 20 − 3 ) ( 20 − 2 ) ( 20 − 3 ) by the FOIL method.
• ⓑ Multiply 18 · 17 18 · 17 without using a calculator.

Writing Exercises

Which method do you prefer to use when multiplying two binomials: the Distributive Property, the FOIL method, or the Vertical Method? Why?

Which method do you prefer to use when multiplying a trinomial by a binomial: the Distributive Property or the Vertical Method? Why?

Multiply the following:

( x + 2 ) ( x − 2 ) ( y + 7 ) ( y − 7 ) ( w + 5 ) ( w − 5 ) ( x + 2 ) ( x − 2 ) ( y + 7 ) ( y − 7 ) ( w + 5 ) ( w − 5 )

Explain the pattern that you see in your answers.

( m − 3 ) ( m + 3 ) ( n − 10 ) ( n + 10 ) ( p − 8 ) ( p + 8 ) ( m − 3 ) ( m + 3 ) ( n − 10 ) ( n + 10 ) ( p − 8 ) ( p + 8 )

( p + 3 ) ( p + 3 ) ( q + 6 ) ( q + 6 ) ( r + 1 ) ( r + 1 ) ( p + 3 ) ( p + 3 ) ( q + 6 ) ( q + 6 ) ( r + 1 ) ( r + 1 )

( x − 4 ) ( x − 4 ) ( y − 1 ) ( y − 1 ) ( z − 7 ) ( z − 7 ) ( x − 4 ) ( x − 4 ) ( y − 1 ) ( y − 1 ) ( z − 7 ) ( z − 7 )

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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• Book title: Elementary Algebra
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Multiplying Polynomials

Multiplication is one of the arithmetic operations which can be applied to polynomials. Multiplying polynomials is one of the simplest things in algebra. Polynomials can be easily multiplied by using their rules. When multiplying polynomials we multiply coefficients together and variables together. In this chapter, we will discuss the multiplication of polynomials, their rules, and the steps to multiply polynomials.

Multiplication of Polynomials

Polynomial multiplication is a method for multiplying two or more polynomials together. The terms of the 1 st polynomial are multiplied with the 2 nd polynomial to get the resultant polynomial. Based on the types of polynomials we use, there are different ways of multiplying them. The rules for the multiplication of polynomials are different for each type of polynomial . To multiply polynomials, the coefficient is multiplied with a coefficient, and the variable is multiplied with a variable.

Multiplying Polynomials with Exponents

When the polynomials are multiplied it is possible they can be monomial, binomial, or trinomial. In order to multiply any two polynomials the steps used are:

• Multiply the coefficients
• Multiply the variables using exponent rules as per the requirement.

Let us understand how to multiply polynomials with exponents using an example.

Example: Multiply 2x 3 with 3x 2

We will follow the same procedure for multiplying polynomials with exponents as we had done above.

• Step 1: First we will multiply the coefficients i.e., 2 × 3 = 6
• Step 2: Next, we will multiply the variables but in this case, the powers of both the variables will be added as per the rules of exponents i.e., x 3 × x 2 = x 5

The final answer is 2x 3 × 3x 2 = 6x 5

Multiplying Polynomials with Different Variables

It is possible to multiply polynomials with different variables too. The steps to multiply polynomials with different variables are:

• Multiply the variables and use rules of exponents wherever necessary.

Example: Multiply 5x 2 with 3y.

• Step 1: We will first multiply the coefficients of both the polynomials i.e., 5 × 3= 15
• Step 2: Since the above polynomials have two different variables, they cannot be multiplied. Hence, we will keep them the same.

The final answer is 5x 2 × 3y = 15x 2 y

How to Multiply Monomials?

Monomials are polynomials having just one term, consisting of a variable and its coefficient. Hence the steps to determine the product of two or three monomials follow the same steps as we learned above. It is possible to multiply monomials more than three too using the same steps we will learn for the below examples.

Multiplication of Two Monomials

When multiplying monomials, we need to follow certain rules similar to multiplying polynomials. Let us understand by taking two monomials, 3x and 2x.

• Step 1 : In the above monomials, the common variable is x. We will multiply the variable with the variable. Hence, we get x × x = x 2 .
• Step 2 : In the next step, we will multiply the coefficients of both the monomials to get 2 × 3 = 6. Thus, multiplying the polynomials 2x and 3x gives 6x 2 as the result.

Multiplication of Three Monomials

To multiply three monomials, we will use the same method as that used for multiplying two monomials. Let us understand the method with an example.

Example: Multiply 2x, 3y, and 6z.

• Step 1 : First we will multiply the variables together i.e., x × y × z = xyz
• Step 2 : Next we will multiply the coefficients of all the three terms i.e., 2 × 3 × 6 = 36

Thus, the multiplication result can be shown as 2x × 3y × 6z = 2 × 3 × 6 × x × y × z = 36xyz

How to Multiply Binomials?

Binomials are a particular kind of polynomials consisting of only two terms. They can be multiplied in two ways:

• Distributive Property

Multiplying Binomials by Distributive Property

For multiplying binomials, we use the distributive property. Let's multiply a binomial (a+b) with another binomial (c+d).

• Step 1: Write both the binomials together i.e., (a + b)(c + d)
• Step 2: Out of the two brackets, keep one bracket constant, let's say (c + d).
• Step 3: Now multiply each and every term from the other bracket i.e., (a + b) with (c + d).

Multiply (2x+3)(4x+5)

The above polynomials can be solved as: (2x + 3)(4x + 5) = 2x(4x + 5) + 3(4x + 5) ⇒ 8x 2 + 10x + 12x + 15 ⇒ 8x 2 + 22x + 15

Multiplying Binomials by Box Method

Two binomials can also be multiplied using the box method. The terms are written across a box and their corresponding products are written inside the box.

Multiply (x + 7) with (x + 3) Solution: Let's write the polynomials (x + 7) horizontally and (x + 3) vertically. Take the sign with its corresponding term on the right. After multiplying the corresponding terms, we get:

Thus, the above multiplication method is known as the box multiplication of two binomials. We now have (x 2 + 7x + 3x + 21) as the sum. Thus, the final product will be (x 2 + 10x + 21).

How to Multiply a Monomial with a Binomial?

As we did above, to multiply a monomial with a binomial, we have to use the distributive property. Let's say monomial a has to be multiplied with binomial (b + c). By distributive property, the above product can be written as: a(b + c) = ab + ac.

Multiply 3y with (5x + 2z)

3y(5x + 2z) = 3y × 5x + 3y × 2z

⇒ (3 × 5 × y × x) + (3 × 2 × y × z) = 15yx + 6yz

Topics Related to Multiplying Polynomials

Check out these interesting articles to learn more about multiplying polynomial and its related topics.

• Multiplying Polynomials Calculator
• Multiplying Binomials Calculator
• Polynomial Calculator

Tips to Remember

• When multiplying polynomials, the coefficient will be multiplied with a coefficient and the variable will be multiplied with a variable.
• Polynomials can also be solved using the distributive property, box method, or grid method.
• When multiplying polynomials with exponents, the rules of exponents have to be used.

Multiplying Polynomials Examples

Example 1: Simplify xz(x 2 + z 2 ) by using rules of multiplication of polynomials.

Solution: xz(x 2 + z 2 ) = (xz × x 2 ) + (xz × z 2 )

⇒ x 3 z + xz 3

Therefore, the product is x 3 z + xz 3

Example 2: Find the product: (2x + 3y)(4x - 5y)

Solution: By using distributive property for multiplying polynomials, we get

2x(4x -5y) + 3y(4x - 5y) = 8x 2 - 10xy + 12xy -15y 2

⇒ 8x 2 + 2xy - 15y 2

Therefore, the product is 8x 2 + 2xy - 15y 2

Example 3: A cuboid has sides measuring 2y, 3x and 5z as its length, breadth, and height respectively. Find the volume of the cuboid.

Solution: As we know, the volume of cuboid = length × breadth × height. Here, all the side lengths are given in the form of monomials, so by applying rules for multiplying monomials, we get,

⇒ Volume = 2y × 3x × 5z = 30xyz

Therefore, the volume of the cuboid is 30xyz cubic units.

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Practice Questions on Multiplying Polynomials

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FAQs on Multiplying Polynomials

How do you multiply three polynomials.

Multiplication of three polynomials is a two-step process that involves the following two steps:

• Multiplication of coefficients
• Multiplication of the variables using Laws of Exponents as and when required.

Let's take an example to understand the multiplication of three polynomials. Example: Multiply (3m+2), 4n 2 , and 7p.

• The above given three polynomials are written as (3m+2)× 4n 2 × 7p
• By using distributive property of polynomial multiplication we get, ((3m× 4n 2 )+(2× 4n 2 ))× 7p = (12mn 2 + 8n 2 )7p = 84mn 2 p + 56n 2 p

Thus, the above multiplication can be shown as (3m+2)× 4n 2 × 7p = 84mn 2 p + 56n 2 p.

How can we Multiply Polynomials Using the Box Method?

Two or more polynomials can be multiplied using the box method. The terms are written across a box and their corresponding products are written within the box. Example: (3x 2 + 2x + 4)(4x + 5)

3x 2 +2x+4 will be written on the vertical side of the box while 4x+5 will be written on the horizontal side of the box, or vice-versa. Then, first, we will multiply 3x 2 by 4x, then 3x 2 by 5, and write the products in the corresponding section of the box. Secondly, we will multiply 2x by 4x and 2x by 5 and write down the products. The final column of the box is filled by multiplying 4 by 4x and 4 by 5. At last, we will add all six terms obtained to get the final answer. Therefore, the result of the multiplication of both the polynomials is (12x 3 +23x 2 +26x+20).

How do you Multiply Binomials Using the Grid Method?

The steps to multiply polynomials by a box method or the grid method is as follows: Example: (x + 6)(2x + 3)

x+6 will be written on the vertical side of the box while 2x+3 will be written on the horizontal side of the box, or vice-versa. Multiply each term with the respective terms. Therefore, the product which we get is (2x 2 + 15x + 18).

How Many Methods are there for Multiplying Polynomials?

There are two methods for multiplying polynomials:

What does FOIL Stand for in Multiplying Binomials?

FOIL stands for First, Outer, Inner Last in multiplying binomials. The binomials are multiplied as:

• Step 1: Multiply the first term of each binomial.
• Step 2: Now multiply the outer term of each binomial.
• Step 3: Once this is done, now multiply the inner terms of the binomials.
• Step 4: Now the last terms are multiplied.
• Step 5: Once all the above four steps are done, the products obtained as each step are added, like terms are combined and the answer is simplified.

What is the Best Method for Multiplying Polynomials?

The best method for multiplying polynomials is the distributive property of multiplying polynomials. The steps to multiply a polynomial using the distributive property are:

• Step 1: Write both the polynomials together.
• Step 2: Out of the two brackets, keep one bracket constant.
• Step 3: Now multiply each and every term from the other bracket.

How Do You Multiply Two Trinomials Together?

Two trinomials can be multiplied together by using the box method as well as the distributive property. Let's take an example to understand the multiplication of two trinomials.

Example: Multiply (5xy + 2x + 3) with (x 2 + 3xy + 7)

• The above given two trinomials are written as (5xy + 2x + 3)× (x 2 + 3xy + 7)
• By using distributive property of polynomial multiplication we get, (5xy + 2x + 3)× (x 2 + 3xy + 7) = 5x 3 y + 15x 2 y 2 + 2x 3 + 6x 2 y + 44xy+ 3x 2 + 14x + 21

Thus, the above multiplication can be shown as (5xy + 2x + 3)× (x 2 + 3xy + 7) = 5x 3 y + 15x 2 y 2 + 2x 3 + 6x 2 y + 44xy + 3x 2 + 14x + 21.

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Video transcript

Multiplying Polynomials by Polynomials

Examples, practice problems and steps!

Worksheet on this topic Multiply Polynomial Calc

Let's multiply the polynomial $$(3x^6 + 2x^5 + 5)$$ by the polynomial (5x + 2) .

Distribute .

Add the resulting Polynomials

$$( 15x^7 + 10x^6 + 25x) + (6x^6 + 4x^5 + 10) = 15x^7 + 16x^6 + 4x^6 + 25x + 10 $$

$$( 15x^7 + 10x^6 + 25x) + (6x^6 + 4x^5 + 10) \\ = 15x^7 + 16x^6 + 4x^6 + 25x + 10 $$

Let's multiply the polynomial $$(3x^6 + 2x^5 + 5)$$ by $$(4x^2 + x + 5)$$ .

Add the resulting Polynomials .

$$ (12x^8 + 8x^7 + 20x^2) + (3x^7 + 2x^6 + 5x) + (15x7^6+ 10x^5 + 25) \\ = \boxed{ 12x^8 + 11x^7 + 17x^6 + 10x^5 + 20x^2 + 5x + 25} $$

Practice Problems

Multiply the polynomial (3x + 4) by the polynomial (5x 4 + 7x 3 + 5x) .

This is similar to example 1 .

15x 5 + 41x 4 + 28x 3 + 15x 2 + 20x

Multiply (2x 7 + 4x 2 + 3x)(3x 8 + 2x 3 + 15x) .

This is similar to example 2 .

6x 13 + 4x 12 + 12x 8 + 27x 7 + 6x 6 + 20x 2 + 15x

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Multiplying Polynomials: Definition, Steps, Methods, Examples, Facts

What is the multiplication of polynomials, rules for multiplying polynomials, multiplying polynomials with different or mixed variables, solved examples on multiplying polynomials, practice problems on multiplying polynomials, frequently asked questions on multiplying polynomials.

To multiply two polynomials, we use the distributive property of multiplication where we multiply each term of the first polynomial by each term of the second polynomial. While doing so, we multiply variables with variables, coefficients with coefficients, and finally add all the results together.

There are different types of polynomials based on the number of terms, such as monomials, binomials, trinomials, and so on. We use different strategies to multiply different types of polynomials together. 

In the multiplication of polynomials, the polynomials can be monomials, binomials, trinomials, etc. The steps to multiply two or more polynomials are as follows:

Step 1: If any one of the polynomials contains more than one term, then first apply the distributive law of multiplication to multiply each term in one polynomial by each term in the other polynomial.

Step 2: Using the laws of exponents, add the powers of the same variables.

Step 3: Add or subtract the like terms and then simplify to get the final results.

Related Worksheets

Multiplying Polynomials using Exponents Law

When we multiply two terms in which the same variables are present, we use the laws of exponents.

a m  × a n = a m + n

Note that we multiply the coefficients of the variables separately. 

Example: $11 x^{2} \times 2x^{5} = 22x^{7}$

It is not necessary that the polynomials being multiplied should be in the same variable. If the two polynomials have different or mixed variables, we use the following rules.

• Multiply all the coefficients. 
• Apply the exponent rules when multiplying the same variables.
• Other variables are simply multiplied together in the answer.

$4x^{2}\;y \times 3yz = 12x^{2}\;y^{2}\;z$

$2x \times (\;-\; 5)xy \times 7z = \;-\; 70x^{2}\;yz$

How To Multiply Polynomials

To learn how to multiply polynomials, we will discuss different categories, such as

• Multiplying monomials
• Multiplying binomials
• Multiplying a monomial and a binomial
• Multiplying a binomial and a trinomial

This way it will be easier to understand how to approach different types of problems involving multiplying polynomials. Let’s discuss these cases in detail.

Multiplication of Monomials

Monomial is a polynomial containing only one term. This term can be a constant, a variable with a coefficient or a product of one or more variables with a coefficient.

Multiplication of Two Monomials

Here, we just have to multiply the single term of the first monomial by the single term of the other monomial. Multiply variables with variables, coefficients with coefficients, and combine the result. If both the monomials have the same variable then their exponents are added together.

Examples:  

$4x \times 5x = (4 \times 5) \times (x \times x) = 20x^{1+1} = 20x^{2}$

$8xy \times 7xy^{2} = (8 \times 7) \times (x \times x) \times (y \times y^{2}) = 56\; x^{2}\;y^{3}$

Multiplication of Three Monomials

Here, we multiply three terms together. We can multiply three monomials by applying the same method as multiplying two monomials.

$4x \times 5x \times 6x = (4 \times 5 \times 6) \times (x \times x \times x) = 120x^{1+1+1} = 120x^{3}$

$x \times 6xy \times xyz = (1 \times 6)(x \times x \times x) \times z = 6x^{3}\;y^{2}\;z$

Multiplication of Binomials

A binomial is a polynomial with only two terms. To multiply binomials, we use the distributive property of multiplication. We first multiply the first term of the first binomial by each term of the second binomial. Next, we multiply the second term of the first binomial by each term of the second binomial. Finally, we combine all the results using the operators.

The mnemonic FOIL can be used to remember and understand the pattern used in the multiplying binomials.

F: First terms 

O: Outer terms 

I: Inner terms 

L: Last terms

Multiplying Binomials Using Distributive Property

Multiplication of a binomial $(a + b)$ with a binomial $(c + d)$ can be given as

$(a + b) \times (c + d) = a \times (c + d) + b \times (c + d)$

          $= ac + ad + bc + bd$

$(x \;-\; 2y ) \times (x^{3} + yz) = x(x^{3} + yz) \;-\; 2y(x^{3} + yz)$

$ = x^{4} + xyz \;-\; 2x^{3}\;y\;-\;2y^{2}\;z$

Multiplying Binomials Using the Box Method

We can multiply two binomials using the box method. Write the individual terms of two binomials along the rows and columns of a 22 box. Every cell contains the product of the terms it represents along the row and the column. Add all the products together to get the final result.

Example: ( x – 2) (y+xz)

$(x \;-\; 2)(y + xz) = x^{2}\;z + xy \;-\; 2xz \;-\; 2y$

How to Multiply a Monomial with a Binomial

To multiply a monomial with a binomial use the distributive property of multiplication.

Multiplication of a monomial a with a binomial $(b + c)$ can be given as

$a(b + c) = (a \times b) + (a \times c)$

$2x \times (x + y) = 2x \times x + 2x \times y = 2x^{2} + 2xy$

How to Multiply Binomial with a Trinomial

A binomial (m + n ) can be multiplied with a trinomial $(a + b + c)$ as

$(m + n) \times (a + b + c) = \left[m (a + b + c)\right] + \left[n (a + b + c)\right]$

       $= am + bm + cm +  an + bn + cn$

$2xy \times (x + 5y + z) = 2x^{2}\;y + 2xyz$

Polynomial Long Multiplication

The horizontal multiplication method is convenient to use when we are dealing with small polynomials. However, if we wish to multiply polynomials having three or more terms, we can prefer the long multiplication method. It is exactly the same as the method of long multiplication for numbers.

Let’s understand with an example.

Example: $(y^{2} + 2y \;-\; 1) \times (y + 5)$

Facts about Multiplying Polynomials

• Multiplication of polynomials follows commutative property as well as associative property .
• The degree of the product of two or more polynomials with one variable is less than or equal to the sum of the degrees of each polynomial.
• A variable without exponent in a polynomial is said to have exponent 1. $x = x^{1}$
• If you multiply a polynomial with a non-zero constant, the degree of the polynomial will not be affected.
• Multiplying polynomials requires careful attention to the signs of the terms.

In this article, we have discussed the multiplication of polynomials with the help of different cases. Let’s solve a few examples and practice problems to recall and revise these concepts.

1. Find the product of $4x$ and $9x$ .

We have to multiply two monomials $4x$ and $9x$.

Required product $=4x \times 9x$

     $=(4 \times 9)x^{1+1}$

     $=36x^{2}$

2. Find the product of $p$ , $3p$ , and $5p + q$ .

$p \times 3p \times (5p + q) = 3 \times 5 \times p \times p \times (5p + q)$

        $=15p^{2} \times (5p + q)$

        $= (15p^{2} \times 5p) + (15p^{2} + q)$

         $= 75p^{3} + 15p^{2} \;q$

3. Multiply the two binomials: $(x^{2} + 7x)(x + 1)$ .

Verify the answer using the box method.

$(x^{2} + 7x)(x + 1) = x^{2}(x + 1) + 7x (x + 1)$

                 $= x^{3} + x^{2} + 7x^{2} + 7x$

                 $= x^{3} + 8×2 + 7x$

Box method:

$\bigg(x^{2} + 7x \bigg)(x +1) = x^{3} + x^{2} + 7x^{2} + 7x = x^{3} + 8x^{2}$

4. Find the product of y and $(3y + 2)$ .

Let’s multiply the monomial y and binomial $(3y + 2)$.

Required product $= y(3y + 2) = (3y \times y) + 2y = 3y^{2} + 2y$

5. Find the product of $(x + 1)$ and $(x + 4)$ .

Solution: 

Required product $= (x + 1)(x + 4)$

     $=x (x + 4) + 1(x + 4)$

     $=x \times x + 4x + x + 4$

     $= x^{2} + 5x + 4$

6. Multiply $(2x^{2} + 5)(x^{2} \;-\; 3x + 2)$ using long multiplication.

Attend this quiz & Test your knowledge.

What is the product of $4t^{2}$ and $7^{t}$?

What is the product of x and $(x \;–\; 3)$, what is the coefficient of the product of $5x^{2},\; 4y$, and $2y$, what is the product of $(2x \;–\; 1)(3x + 4)$, the product of $(l + 3m)$ and $(l\;-\;2m)$ is equal to:.

What is binomial?

A polynomial that has only two terms is called a binomial.

What is trinomial?

A polynomial that has only three terms is called a trinomial.

How do you multiply polynomials with mixed variables?

Multiply all the coefficients. Multiply the same variables using the laws of exponents. Multiply these two products with the remaining variables to get the answer.Example: $5x \times 2xy \times 5xyz = 50x^{3}\;y^{2}\;z$

What is the product of two or more monomials?

The product of two or more monomials is also a monomial.

What is the degree of the product polynomial when multiplying two polynomials of the same degree?

When multiplying two polynomials of the same degree, the degree of the product polynomial is twice the degree of the two polynomials being multiplied.

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Multiplying Polynomials

Learning outcome.

• Multiply polynomials

Another type of polynomial multiplication problem is the product of a binomial and trinomial. Although the FOIL method cannot be used since there are more than two terms in a trinomial, you still use the Distributive Property and the Vertical Method to organize the individual products. Using the distributive property, each term in the binomial must be multiplied by each of the terms in the trinomial. The most important part of the process is finding a way to organize terms.

Multiply using the Distributive Property: [latex]\left(x+3\right)\left(2{x}^{2}-5x+8\right)[/latex]

Find the product.  [latex]\left(3x+6\right)\left(5x^{2}+3x+10\right)[/latex]

Distribute the trinomial to each term in the binomial.

[latex]3x\left(5x^{2}+3x+10\right)+6\left(5x^{2}+3x+10\right)[/latex]

Use the distributive property to distribute the monomials to each term in the trinomials.

[latex]3x\left(5x^{2}\right)+3x\left(3x\right)+3x\left(10\right)+6\left(5x^{2}\right)+6\left(3x\right)+6\left(10\right)[/latex]

[latex]15x^{3}+9x^{2}+30x+30x^{2}+18x+60[/latex]

Group like terms.

[latex]15x^{3}+\left(9x^{2}+30x^{2}\right)+\left(30x+18x\right)+60[/latex]

Combine like terms.

[latex]\left(3x+6\right)\left(5x^{2}+3x+10\right)=15x^{3}+39x^{2}+48x+60[/latex]

As you can see, multiplying a binomial by a trinomial leads to a lot of individual terms! Using the same problems as above, we will use the vertical method to organize all the terms produced by multiplying two polynomials with more than two terms.

Multiply using the Vertical Method: [latex]\left(x+3\right)\left(2{x}^{2}-5x+8\right)[/latex]

Solution It is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products this way.

Multiply. [latex]\left(3x+6\right)\left(5x^{2}+3x+10\right)[/latex]

Set up the problem in a vertical form, and begin by multiplying [latex]3x+6[/latex] by [latex]+10[/latex]. Place the products underneath, as shown.

[latex]\begin{array}{r}3x+\,\,\,6\,\\\underline{\times\,\,\,\,\,\,5x^{2}+\,\,3x+10}\\+30x+60\,\end{array}[/latex]

Now multiply [latex]3x+6[/latex] by [latex]+3x[/latex]. Notice that [latex]\left(6\right)\left(3x\right)=18x[/latex]; since this term is like [latex]30x[/latex], place it directly beneath it.

[latex]\begin{array}{r}3x\,\,\,\,\,\,+\,\,\,6\,\,\\\underline{\times\,\,\,\,\,\,5x^{2}\,\,\,\,\,\,+3x\,\,\,\,\,\,+10}\\+30x\,\,\,\,\,+60\,\,\\+9x^{2}\,\,\,+18x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Finally, multiply [latex]3x+6[/latex] by [latex]5x^{2}[/latex]. Notice that [latex]30x^{2}[/latex] is placed underneath [latex]9x^{2}[/latex].

[latex]\begin{array}{r}3x\,\,\,\,\,\,+\,\,\,6\,\,\\\underline{\times\,\,\,\,\,\,5x^{2}\,\,\,\,\,\,+3x\,\,\,\,\,\,+10}\\+30x\,\,\,\,\,+60\,\,\\+9x^{2}\,\,\,+18x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+15x^{3}+30x^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}[/latex]

Now add like terms.

[latex]\begin{array}{r}3x\,\,\,\,\,\,+\,\,\,6\,\,\\\underline{\times\,\,\,\,\,\,5x^{2}\,\,\,\,\,\,+3x\,\,\,\,\,\,+10}\\+30x\,\,\,\,\,+60\,\,\\+9x^{2}\,\,\,+18x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+15x^{3}\,\,\,\,\,\,+30x^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\+15x^{3}\,\,\,\,\,\,+39x^{2}\,\,\,\,+48x\,\,\,\,\,+60\end{array}[/latex]

The answer is [latex]15x^{3}+39x^{2}+48x+60[/latex].

Notice that although the two problems were solved using different strategies, the product is the same. Both the horizontal and vertical methods apply the distributive property to multiply a binomial by a trinomial.

In our next example, we will multiply a binomial and a trinomial that contains subtraction. Pay attention to the signs on the terms. Forgetting a negative sign is the easiest mistake to make in this case.

Find the product.

[latex]\left(2x+1\right)\left(3{x}^{2}-x+4\right)[/latex]

[latex]\begin{array}{cc}2x\left(3{x}^{2}-x+4\right)+1\left(3{x}^{2}-x+4\right) \hfill & \text{Use the distributive property}.\hfill \\ \left(6{x}^{3}-2{x}^{2}+8x\right)+\left(3{x}^{2}-x+4\right)\hfill & \text{Multiply}.\hfill \\ 6{x}^{3}+\left(-2{x}^{2}+3{x}^{2}\right)+\left(8x-x\right)+4\hfill & \text{Combine like terms}.\hfill \\ 6{x}^{3}+{x}^{2}+7x+4 \hfill & \text{Simplify}.\hfill \end{array}[/latex]

Another way to keep track of all the terms involved in the above product is to use a table, as shown below. Write one polynomial across the top and the other down the side. For each box in the table, multiply the term for that row by the term for that column. Then add all of the terms together, combine like terms, and simplify. Notice how we kept the sign on each term; for example, we are subtracting [latex]x[/latex] from [latex]3x^2[/latex], so we place [latex]-x[/latex] in the table.

Multiply.  [latex]\left(2p-1\right)\left(3p^{2}-3p+1\right)[/latex]

Distribute [latex]2p[/latex] and [latex]-1[/latex] to each term in the trinomial.

[latex]2p\left(3p^{2}-3p+1\right)-1\left(3p^{2}-3p+1\right)[/latex]

[latex]2p\left(3p^{2}\right)+2p\left(-3p\right)+2p\left(1\right)-1\left(3p^{2}\right)-1\left(-3p\right)-1\left(1\right)[/latex]

Multiply. Notice that the subtracted [latex]1[/latex] and the subtracted [latex]3p[/latex] have a positive product that is added.

[latex]6p^{3}-6p^{2}+2p-3p^{2}+3p-1[/latex]

[latex]6p^{3}-9p^{2}+5p-1[/latex]

In the following video, we show more examples of multiplying polynomials.

Multiplication of binomials and polynomials requires use of the distributive property as well as the commutative and associative properties of multiplication. Whether the polynomials are monomials, binomials, or trinomials, carefully multiply each term in one polynomial by each term in the other polynomial. Be careful to watch the addition and subtraction signs and negative coefficients. A product is written in simplified form if all of its like terms have been combined.

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• Published: 12 April 2024

An efficient polynomial-based verifiable computation scheme on multi-source outsourced data

• Yiran Zhang 1   na1 ,
• Huizheng Geng 1   na1 ,
• Shen He 1 &

Scientific Reports volume  14 , Article number:  8512 ( 2024 ) Cite this article

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• Computer science
• Information technology

With the development of cloud computing, users are more inclined to outsource complex computing tasks to cloud servers with strong computing capacity, and the cloud returns the final calculation results. However, the cloud is not completely trustworthy, which may leak the data of user and even return incorrect calculations on purpose. Therefore, it is important to verify the results of computing tasks without revealing the privacy of the users. Among all the computing tasks, the polynomial calculation is widely used in information security, linear algebra, signal processing and other fields. Most existing polynomial-based verifiable computation schemes require that the input of the polynomial function must come from a single data source, which means that the data must be signed by a single user. However, the input of the polynomial may come from multiple users in the practical application. In order to solve this problem, the researchers have proposed some schemes for multi-source outsourced data, but these schemes have the common problem of low efficiency. To improve the efficiency, this paper proposes an efficient polynomial-based verifiable computation scheme on multi-source outsourced data. We optimize the polynomials using Horner’s method to increase the speed of verification, in which the addition gate and the multiplication gate can be interleaved to represent the polynomial function. In order to adapt to this structure, we design the corresponding homomorphic verification tag, so that the input of the polynomial can come from multiple data sources. We prove the correctness and rationality of the scheme, and carry out numerical analysis and evaluation research to verify the efficiency of the scheme. The experimental indicate that data contributors can sign 1000 new data in merely 2 s, while the verification of a delegated polynomial function with a power of 100 requires only 18 ms. These results confirm that the proposed scheme is better than the existing scheme.

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Introduction

Cloud computing technology has become an indispensable tool in the Internet era, and it has gradually penetrated into all aspects of daily life. Users with weak computing capabilities tend to outsource complex computing tasks to cloud servers with powerful computing and storage capabilities, thus reducing the complexity of local computing 1 . However, cloud servers are not completely trustworthy. The cloud servers have the potential to leak user data or intentionally return incorrect calculation result. Therefore, it is of practical significance to verify the calculation results of outsourcing services without disclosing user privacy 2 , 3 .

Verifiable computation (VC, verifiable computation) solves the above problem. The user sends the function and the input data of the function to the cloud server, and the cloud server returns the calculated result and the proof of the result. Users can verify the correctness of the calculation results, and the computational complexity of this process is much smaller than that of directly calculating functions. Verifiable computation is generally divided into two categories: (1) verifiable computation of general functions, which is suitable for the computation of any function 4 , 5 ; (2) verifiable computation of special functions, such as modular exponential operation 6 , 7 , polynomial calculation 8 , attribute-based decryption operation 9 , etc. Among them, the polynomial-based verifiable computation is widely used in information security, linear algebra, signal processing and other fields, so it has attracted wide attention.

Motivation and contribution

The researchers have proposed some verifiable computation based on polynomial schemes 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 , 33 , 34 , 35 . Benabbas et al. 10 put forward a polynomial outsourcing computing scheme for the first time. The scheme requires that the input of the polynomial must come from a single data source, which means that the data must be signed by a single user. However, the input of the polynomial may come from multiple users in the practical application. In order to solve this problem, the scheme 11 is proposed to design an outsourced polynomial computation program based on the idea of homomorphic verifiable computation tags, and make the scheme support multiple data sources. However, the scheme requires that all multiplication gates must be executed before the addition gate when generating verification tag, which greatly limits the speed of generating the verification tag and leads to low efficiency. Want et al. 12 improved the scheme 11 , but it only improved the security of signatures and do not consider the efficiency degradation caused by he design of verification tag. In particular, when the order of the polynomial function is relatively high and the data of the same user is calculated many times, the correctness verification of the result will be extremely slow.

From the above references, there are two problems with the proposed scheme. First, the existing scheme requires that the input of the polynomial must come from a single data source; Secondly, the design of verification tags may cause a decrease in efficiency, especially when the polynomial function is relatively complex, so that the result correctness verification process will be extremely slow, and even affect the use of data. Therefore, we define two key requirements for efficient verifiable computation schemes on multi-source outsourced data. (1) Efficient. The scheme should ensure that the verification can be completed quickly. (2) Support for multiple data sources. The input of the polynomial can come from multiple independent data sources, which means that the data from different data sources can be signed with different private keys.

To address these issues, we propose a new and efficient polynomial-based verifiable computation scheme on multi-source outsourced data, which has the characteristics of efficient and supporting multiple data sources. We optimize the polynomials using Horner’s Method to increase the speed of verification, in which the addition gate and the multiplication gate can be interleaved to represent the polynomial function. In order to adapt to this structure, we design the corresponding verification tag, which is additive homomorphism and multiplicative homomorphic, so as to suit for all types of polynomials. We have verified the correctness and soundness of the scheme based on Computational Diffie-Hellman(CDH) Assumption. The experimental prove the efficiency of the scheme.

The main contributions of this paper can be summarized as follows:

We design for the first time an efficient polynomial-based verifiable computation scheme on multi-source outsourced data, which has the characteristics of efficient and supporting multiple data sources. For multi-source outsourcing systems, the cloud server can perform polynomial functions to obtain the calculation results and generate proof information, which can be used by third parties to verify the correctness of the calculation results without knowing the input.

In order to solve the problem of single data source, this paper designs a homomorphic verification tag structure that supports multiple data sources. As the polynomial function is executed gate by gate, we use the key management center to convert the signatures signed by different user into the verification tag with the unified public and private keys, so that the input of the polynomial can come from multiple data sources.

In order to solve the problem of low efficiency, we optimize the polynomials using Horner’s Method, and the generation of corresponding verification tag can be generated with the cross-operator of multiplication gate and addition gate, so as to improve the verification speed.

Related work

Gennaro et al. 13 combined outsourced computation and verification technology to propose the concept of verifiable computation for the first time. It constructed an outsourced scheme of verifiable computation by using obfuscated circuits and full homomorphic encryption, which can ensure the privacy of input and output. However, this scheme can only do private verification. Benabbas et al. 10 proposed a polynomial outsourcing computing scheme with Chosen Plaintext Attack (CPA) security, which solved the problem left by Gennaro et al. 13 The scheme used addition homomorphic encryption algorithm to ensure the privacy of the polynomial, but could not guarantee the privacy of inputs and realize public verification. Zhang et al. 14 constructed a univariate polynomial outsourcing calculation scheme by using multilinear mapping and homomorphic encryption algorithm. This scheme can ensure the privacy of input, and its extension scheme can ensure the privacy of function, but it can only achieve private verification. Papamanthou et al. 15 proposed a verifiable outsourcing computation scheme for dynamic polynomials that allows incremental updating of the coefficients. Fiore et al. 16 proposed a verifiable polynomial outsourcing computation scheme with adaptive security, but this scheme can only guarantee the privacy of the function. Zhang et al. 18 improved the efficiency of IOT cross-chain computing by outsourcing polynomials to the blockchain, and they proposed an efficient and verifiable polynomial cross-chain outsourcing computing scheme for verifying the correctness of the results of calculations on the blockchain, but the practicality of the scheme is modest.

Other researchers have proposed verifiable computation schemes based on homomorphic signatures 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 . Barbosa et al. 19 put forward the Delegatable Homomorphic Encryption (DHE) cryptographic primitive, and give a method on how to use DHE to construct a verifiable computation scheme. In recent years, Guo et al. 20 has developed a lightweight verifiable blind decryption technique based on a linear homomorphic encryption scheme to verify the correctness of the final result. Boneh and Freeman 21 proposed the implementation of homomorphic signatures based on polynomials of constant degree, but this scheme can only be applied to the verifiable computation of polynomials of constant degree. Fiore and Gennaro 22 proposed a publicly verifiable secure outsourcing protocol for polynomial and matrix multiplication evaluation. However, this scenario does not support multiple data contributors. Song et al. 23 proposed a verifiable computation scheme that supports multiple data sources. It is based on the verifying data structure of the Homomorphic Verifiable Computation Tags, which is only an additive homomorphism. However, polynomials not only have addition operations, but also multiplication operations, so the scheme is not functional enough to support polynomial computation. Further, song et al. 11 proposed a verifiable polynomial computation scheme that supports multiple data sources. When two inputs are signed by different keys from different data contributors, it is difficult to have a uniform validation data structure to support addition and multiplication. To solve this problem, the idea is to place all addition gates behind product gates to represent delegated polynomial functions. Then, based on this structure, they further designed the first-level verification label and the second-level verification label. By utilizing these designs, the server is able to output homomorphic validation labels for each gate even if the validation labels for the two inputs are signed by different keys. However, it is clear that executing the addition gate after the multiplication gate will affect the speed of the verification label generation, especially if the input with a data source is evaluated multiple times. Although Want et al. 12 improved the scheme 11 , it only improved the security of signatures and did not pay attention to the inefficiency. Although there are few solutions solve the problem of correctness verification of polynomial calculation with multi-sources, they do not pay attention to the efficiency reduction caused by the designing of scheme.

Preliminaries

Arithmetic circuit.

Arithmetic circuits 36 on fields F and variable sets \(X={x_1,\ldots ,x_n}\) have two kinds of gates: multiplication gate ’ \(\times\) ’ and addition gate ’+’. Every gate marked with the ’x’ is called the product gate, and every gate marked with the ’+’ is called the sum gate.

The arithmetic circuit computes polynomial functions, where the product gate computes the product of polynomials on its input wire, and the summation gate computes the sum of polynomials on its input wire. In this paper, the cloud server performs gate-to-gate processing of polynomials based on arithmetic circuits.

Bilinear mapping

Bilinear mapping refers to the linear mapping relationship between two cyclic groups 37 . We define the mapping \(e:G_1\times G_1\rightarrow G_2\) as a bilinear mapping, where \(G_1\) and \(G_2\) are multiplicative cyclic group of order p , and g ,  h are two generators of the group \(G_1\) . It satisfies the following properties:

Bilinear: for \(a,b\in Z_p, g^a,g^b,h^a,h^b\in G_1\) , then \(e\left( g^a,g^b\right) =e{(g,h)}^{ab}\) can be calculated.

Non-degenerate: \(e\left( g,g\right) \ne 1\) .

Computability: For any \(g,h\in G_1\) , there are effective algorithms that can calculate \(e\left( g,h\right)\) .

Computational Diffie-Hellman (CDH) Assumption: For x , \(y\in Z_p\) , there are g , \(g^x\) , \(g^y\in G_1\) , then it is difficult to compute \(g^{xy}\) .

Horner’s method

Horner’s method 38 is a polynomial evaluation method with a single data source, aiming to simplify polynomial calculation. It transfers a polynomial of degree n to n linear functions of degree one, and it can be represented as an equation:

For a polynomial \(f\left( x\right)\) with a single data source, it only needs to perform n multiplications and n additions, with a time complexity of \(\mathcal {O}\left( n\right)\) . Compared to normal evaluation, which requires \(n(n+1)/2\) multiplications and n additions, resulting in a time complexity of \(\mathcal {O}\left( n^2\right)\) , Horner’s Method is a faster and better way to compute higher-order polynomials.

Problem statement

System model.

There are three entities in the system model of this scheme: the cloud, the users, and the key management center (KMC).

It provides storage services for users and computes polynomial functions on outsourced data. And it generates the proof message to verify the correctness of the calculation results. It’s not entirely trustworthy.

They outsource their data to the cloud. And they also upload signature data to verify the correctness of the polynomial calculation results. We assume that there are \(n(n\ge 1)\) users. They are completely trustworthy.

Key management center

It assigns keys to users and helps the cloud generate verification information. After the polynomial function is computed, it verifies the correctness of the result based on the proof information. It’s completely trustworthy.

Figure 1 represents the system model of “An efficient polynomial-based verifiable computation scheme on multi-source outsourced data”. In a multi-source data verifiable computing system, there are n user \(u_i(1\le i\le n)\) . Each user \(u_i\) holds their own public key and private key. The user \(u_i\) generates signature by signing the data with the private key, then the user \(u_i\) uploads signature and data to the cloud. The cloud calculates the polynomial to obtain the calculation result, and it also outputs the proof information. The cloud sends the result and the proof information to KMC. KMC helps the user verify the correctness of the result using the proof information.

System model.

Threat model

The cloud is not completely trustworthy. It may cause misbehavior due to monetary reasons, hacking or system failure. In practical applications, there is a risk that the cloud server may produce incorrect calculation results for users without actually performing the computation. The cloud server may even deliberately provide incorrect calculations. Consider, for instance, a scenario where 5000 users (acting as urban pollution data collection points) in 5000 cities gather information on air pollution from various locations. These users upload their air pollution data to the cloud daily, and request that the cloud calculate the average air pollution based on data from multiple locations. However, the cloud server may perform the calculation using only a subset of the data instead of the entire dataset, leading to erroneous results. Even worse, the cloud server may not perform the computation and return historical data directly or generate random numerical values for the user. Therefore, the work is primarily motivated by the need to provide a verifiable polynomial evaluation scheme. This scheme allows users to verify that the cloud server has correctly executed the entrusted polynomial function. The security threats of this scheme are as follow:

Data corruption: The adversary may compromise data during the data is uploaded to the cloud. The corrupted data used as input for polynomial may result in incorrect results.

Incorrect results: For monetary reasons, the cloud may not be able to fully execute the entrusted polynomial, or output the result randomly to save computing resources.

Forgery attacks: The adversary may forge the signature and proof information on purpose, in order to trick the user into passing the correctness verification.

Security goal

The security goal of the proposed scheme is twofold: correctness and soundness.

Correctness: the cloud performed the polynomial correctly, then the corresponding proof information can pass the correctness check of result, that is, there are no false negatives.

Soundness: the verification information corresponding to the wrong result must be detected and fail the correctness check, that is, there are no false positives.

Efficient polynomial-based verifiable computation scheme on multi-source outsourced data

Notations in this section.

Table 1 shows some important notations.

The system executes the algorithm SetUp() to initialize the system parameters. KMC performs the algorithm KeyGen() to obtain the public keys and private keys. The users execute algorithm Sign() for signing the data, and the data and the corresponding signatures are outsourced to cloud. The cloud computes the polynomial function to obtain calculation result, and the cloud executes the algorithm GateVal() to obtain the verification tag. As the circuit is executed gate by gate, the verification tag of the last gate is output as the final proof information. The cloud executes the algorithm ProofCre() which sends the proof information to the KMC. KMC executes the algorithm VerifyProof() which verifies the correctness of the final calculation result. If the output is True, it shows the result is correct; if the output is False, it shows that the result is incorrect.

The proposed scheme

\({setup\left( 1^\lambda \right) \rightarrow (e,p,g_1,g_2,h,h)}\).

The algorithm is executed by the cloud to generate system parameters. The input is the security parameter \(\lambda\) , and the output is the security parameter of system \({{e,p,G_1,G_2,h,H}}\) .

Suppose that \(G_1,G_2\) are two p-order prime groups, g ,  h are generators of \(G_1\) , e is a bilinear mapping \(e:G_1\times G_1\rightarrow G_2\) . \(H:\left\{ 0,1\right\} ^*\rightarrow Z_p\) is a hash function that maps any string to an element in \(Z_p\) .

\(KeyGen\left( 1^k\right) \rightarrow (pk,sk)\)

The algorithm is executed by the key management center to generate public keys and private keys. The input is the security parameter k , and the output is public key pk and private key sk .

KMC randomly selects \(a^*\in Z_p\) as the conversion private key. When a new user joins the system, KMC randomly selects a random number \(a\in Z_p\) as the private key sk of the user, generates and stores \(a\prime\) satisfying \(a*a\prime =a^*\) , and outputs public key \(pk=(g^a,h^a,h^\frac{1}{a})\) . KMC sends the sk and pk to the user, then sends the pk to the cloud.

\(\varvec{Sign\left( m,sk\right) \rightarrow \sigma _m}\)

The algorithm is executed by the user to sign the data. The input is the outsourced data m and private key sk , and the output is the signature \(\sigma _m\) .

We set the label \(\tau\) , which is selected by the user to express the physical implication for the data m . And the label \(\tau\) is public. The user computes \(t_\tau =H\left( \tau \right)\) , chooses k at random, computes \(r=h^k,s=h^{a\left( t_\tau +m+k\right) }\ mod\ p\) , where a is the private key. Then it generates signature \(\sigma _m=(r=h^k,s=h^{a\left( t_\tau +m+k\right) }\ mod\ p)\) . Finally, the user uploads data m , the label \(\tau\) , and signature \(\sigma _m=(r,s)\) to the cloud.

After receiving data m and the corresponding signature \(\sigma _m\) , the cloud verifies the signature as shown in Eq. ( 2 ), where \(pk^{(1)}=g^a\) . If the verification is successful, the cloud stores the data m , the label \(\tau\) , signature \(\sigma _m\) , otherwise, the cloud outputs \(\bot\) .

Table 2 shows the userID, labels, data, public keys, and signatures of the users. Each user has a public key, such as the public key corresponding to \(u_i\) is represented as \((g^{a_i},h^{a_i},h^{{\frac{1}{a}}_i})(1\le i\le n)\) . A user can upload multiple data. For example, data \(m_1\) and \(m_2\) are uploaded by \(u_1\) .

\(\varvec{GateVal()\rightarrow \delta }\)

The algorithm is executed by the cloud to generate verification tags. The inputs of a gate could be the original outsourced data, the constant \(c\in Z_p\) , or the output of the previous gate. The output is verification tag \(\delta\) . As the circuit is executed gate by gate, the verification tag of the previous gate output is used as the input for the next gate.

Let \(f\left( x_1,\ldots ,x_n\right)\) be a polynomial function, where \(x_i(1\le i\le n)\) represents the outsourced data. Reference 11 gives the definition of the polynomial function \(f\left( x_1,\ldots ,x_n\right) =\sum _{i=1}^{n}{(c_i*\prod _{j} x_j^{e_j})}\) . Then drawing on the idea of Horner’s method 38 , we further represent the delegate function as Eq. ( 3 ).

where \(c_i\) represents the constant coefficient and \(e_j\) represents the exponent of \(x_j\) . It requires that multiplication gates and addition gates can be interleaved to express the delegated polynomial function as shown in Fig. 2 , which improves the situation that the addition gate must be carried out after the multiplication gate in the scheme 11 , so as to improve the verification efficiency. The cloud runs the polynomial function using the arithmetic circuit.

Polynomial functions are represented by arithmetic circuits. For example, \(f=x_1x_2x_3+3x_1x_4x_5+6{x_1x}_4=x_1(x_2x_3+3x_4\left( x_5+2\right) )\) .

If the gate is a multiplication gate, then

The inputs are the constant \(c\in Z_p\) and the variable x which has the verification tag \(\delta (pk,\sigma \left( r,s\right) )\) . For \(y=x*c\) , the GateVal() algorithm outputs the verification tag \(\delta \prime (pk\prime ,\sigma \prime )\) as Eq. ( 4 ).

The inputs are variable \(x_1\) and \(x_2\) with the verification tag \(\delta _1({pk}_1,\sigma _1(r_1=h^{k_1},s_1=h^{a_1\left( t_{\tau 1}+x_1+k_1\right) }mod\ p))\) and \(\delta _2({pk}_2,\sigma _2(r_2=h^{k_2},s_2=h^{a_2\left( t_{\tau 2}+x_2+k_2\right) }mod\ p))\) respectively. For \(y=x_1*x_2\) , the GateVal() outputs the verification tag \(\delta \prime (pk\prime ,\sigma \prime )\) .

The cloud sends \(\sigma _2,x_2\) to KMC.

KMC verifies the signature \(\sigma _2\) as shown in Eq. ( 2 ). If that fails, output \(\bot\) ; otherwise, KMC randomly selects \(k_2^\prime\) , and uses the \(a_1^\prime\) to generate \(s_2^\prime =a_1^\prime \left( t_{\tau 2}+x_2+k_2^\prime \right)\) , \({{\hat{r}=r}_1}^{t_{\tau 2}+x_2+k_2^\prime }\) , \(h^{k_2^\prime }\) , \(pk^\prime =(g^{a^*},h^{a^*},h^\frac{1}{a^*})\) , where \(a_1 *a_1'=a^*\) , and send them to the cloud.

The cloud computes verification tag \(\sigma \prime =(r^\prime ,s^\prime )\) .

where \(\phi =t_{\tau 1}x_2+t_{\tau 1}k_2^\prime +x_1t_{\tau 2}+x_1k_2^\prime +k_1t_{\tau 2}+k_1x_2+k_1k_2^\prime\) .

If the gate is an additive gate ’+’, then

The inputs are the constant \(c\in Z_p\) and the variable x which has the verification tag \(\delta (pk,\sigma \left( r,s\right) )\) . For \(y=x+c\) , the GateVal() algorithm outputs the verification tag \(\delta \prime (pk\prime ,\sigma \prime )\) as Eq. ( 6 ).

The inputs are variable \(x_1\) and \(x_2\) with the verification tag \(\delta _1({pk}_1,\sigma _1\left( r_1,s_1\right) )\) and \(\delta _2({pk}_2,\sigma _2\left( r_2,s_2\right) )\) respectively. For \(y=x_1+x_2\) , the GateVal() algorithm outputs the verification tag \(\delta \prime (pk\prime ,\sigma \prime )\) .

If \({pk}_1={pk}_2\) , this means that \(\delta _1\) and \(\delta _2\) have the same private key, i.e \(a=a_1=a_2\) .

If \({pk}_1\ne {pk}_2\) , this means that \(\delta _1\) and \(\delta _2\) have different private keys, i.e \(a_1\ne a_2\) .

The cloud sends \(\sigma _1,\sigma _2\) to KMC.

KMC generates \(s_1^\prime ={(s_1)}^{a_1^\prime }=h^{a^*\left( t_{\tau 1}+x_1+k_1\right) }\) using \(a_1^\prime\) , generates \(s_2^\prime ={(s_2)}^{a_2^\prime }=h^{a^*\left( t_{\tau 2}+x_2+k_2\right) }\) using \(a_2^\prime\) , and send them to the cloud.

\(ProofCre\left( \delta \right) \rightarrow (P)\)

The algorithm is executed by the cloud to generate the final proof message. The input is the verification tag by running the GateVal() on the last gate and the output is the final proof message \(P=\delta _R(pk,\sigma )\) . The cloud sends the proof message P to KMC.

\(VerifyProof\left( P\right) \rightarrow (True,False)\)

The algorithm is executed by KMC to verify the results of the polynomial calculations. The input is proof message P, and the output is True or False. True shows that the result is correct, False shows that the result is incorrect.

KMC receives the calculation result of the function \(R=f\left( x_1,\ldots ,x_n\right)\) and the proof information \(P=\delta _R(pk,\sigma (r,s))\) . Given that each input \(x_i\) of the polynomial has a label \(\tau _i\) , KMC computes \(t_i=H(\tau _i)\) , then KMC computes \(\rho \leftarrow f\left( t_1,\ldots ,t_n\right)\) . The correctness of the result R is verified using P . If the check is passed, the result R is correct and the output is True. Otherwise, the result R is incorrect and the output is False.

In practice, the data \(\rho \leftarrow f\left( t_1,\ldots ,t_n\right)\) can be generated and stored in advance to increase efficiency.

Security analysis

We analyzed the security of the scheme from two aspects: correctness and soundness. First of all, we confirm that the verification tag designed in this scheme support addition homomorphism and multiplication homomorphism, and on this basis we verify the correctness of the scheme based on the Computational Diffie-Hellman (CDH) Assumption. Secondly, we confirm the soundness of the scheme, in which the verification tag forged by the attacker cannot pass the verification test.

Correctness

We verify that the verification tag designed by the scheme support addition homomorphism and multiplication homomorphism, and then we verify the correctness of the scheme based on CDH hypothesis.

The verification tag is additive homomorphic.

In the addition gate, the inputs \(x_1\) and \(x_2\) have the labels \(\tau _1\) and \(\tau _2\) (for the constant c , the labels are c ), and get \(t_{\tau 1}=H(\tau _1)and t_{\tau 2}=H(\tau _2)\) . For \(y=x_1+x_2\) , the cloud generates verification tags \(\sigma ^\prime =\left( r^\prime ,s^\prime \right) =(h^{k_1+k_2}\) , \(h^{a^*\left( t_{\tau 1}+t_{\tau 2}+(x_1+x_2)+k_1+k_2\right) })\) , where \(a^*\) is the security parameter selected by KMC. Therefore, KMC can verify the correctness of \(y=x_1+x_2\) by Eq. ( 10 ) using the verification tags without knowing \(x_1\) and \(x_2\) .

It is obvious that verification tags are additive homomorphic.

The verifying tag is multiplicative homomorphic .

In the multiplication gate, the inputs \(x_1\) and \(x_2\) have the labels \(\tau _1\) and \(\tau _2\) (for the constant c , the labels are c ), and get \(t_{\tau 1}=H(\tau _1)\) and \(t_{\tau 2}=H(\tau _2)\) . For \(y=x_1*x_2\) , the cloud generates verification tags \(\sigma ^\prime =\left( r^\prime ,s^\prime \right) =(h^k,h^{a^*\left( t_{\tau 1}t_{\tau 2}+x_1x_2+k\right) })\) , where \(a^*\) is the security parameter selected by KMC. Therefore, KMC can verify the correctness of \(y=x_1*x_2\) by Eq. ( 11 ) using the verification tags without knowing \(x_1\) and \(x_2\) .

It is obvious that verification tags are multiplicative homomorphic.

The correctness of the scheme is achieved.

According to Lemmas 1 and 2 , the verification tag of this scheme is a homomorphic verifiable label. KMC can verify the correctness of the calculation results without knowing the input. The correctness of this scheme is equivalent to proofing the correctness of VerifyProof(). The correctness of Eq. ( 9 ) can be verified by Eq. ( 12 ).

The soundness of the scheme is achieved .

We demonstrate the soundness of the scheme, which shows that once the cloud or external attacker is able to pass the verification by forging the verification tag with false result, they are able to establish adversary \(\mathcal {A}\) with a non-negligible probability.

Assume that the security parameters of the system are \((e,p,G_1,G_2,h,H)\) , where \(h=g^x,x\in Z_p\) . Define the proof information \(P=\delta _R(PK,\sigma (r,s)), {PK}^{(1)}=g^a,\) where \(a=x*y(y\in Z_p)\) . The adversary \(\mathcal {A}\) outputs CDH challenge as \((g^{xy},g^y)=({PK}^{(1)},h)\) . Define \(q_{H_i}\) as the number of times to count the process \(t_\tau =H\left( \tau \right)\) . Thus, the probability of the collision occurring in H used to compute \(t_{\tau _i}\) is at most \(q_{H_i}/2^l\) , where l is the length of the output of H . Based on Reset Lemma 39 , adversary \(\mathcal {A}\) can generate two verification tags \(\delta _1(PK,\sigma _1\left( r,s_1\right) ),\ \delta _2(PK,\sigma _2\left( r,s_2\right) )\) with the possibility at least \(\prod _{i=1}^{n}{(\epsilon -\left( \epsilon *q_{H_i}+1\right) /2^l)}^2\) .

The adversary \(\mathcal {A}\) forge verification tags \(\sigma _1(r=h^k,s_1=h^{a\left( \rho _1+R+k\right) }mod\ p),\sigma _2(r=h^k,s_2=h^{a\left( \rho _2+R+k\right) }mod\ p)\) , where \(\rho _1,\rho _2\) are two different outputs of \(\rho \leftarrow f\left( t_1,\ldots ,t_n\right)\) . Adversary \(\mathcal {A}\) can solve the CDH problem by calculating:

It is obvious that if the attacker can forge the verification tag, then we can solve the CDH problem, which is impossible. Therefore, the verifiable computation scheme is soundness, that is, attacker cannot fabricate proof information for any wrong result.

Performance analysis

Communication cost.

In the polynomial verifiable computation scheme, there are two types of communication costs.

The cloud needs to communicate with KMC to transmit proof information. The proof information is expressed as \(P=\delta _R(pk,\sigma )\) , so the communication cost of the proof information is \(\left| S_{pk}\right| +\left| S_\sigma \right|\) , where the size of the public key pk is \(\left| S_{pk}\right|\) and the size of the signature \(\sigma\) is \(\left| S_\sigma \right|\) .

The cloud needs to communicate with KMC to generate corresponding security parameters in GateVal() algorithm. The communication cost of the generated intermediate parameter in the multiplication gate is \(c_**\left| S_G\right|\) , where \(\left| S_G\right|\) indicates the size of the data in \(G_1\) , and \(c_*\) is 0 or 1 indicates the two calculation methods of the multiplication gates. For the additive gate of the polynomial function, the communication cost of the generated intermediate parameter is \(c_+*\left| S_G\right|\) , where \(c_+\) is 0, 1, and 2 indicate the three computation methods of the additive gates. Thus the communication cost is \(\left( c_**{sum}_*\right) *\left| S_G\right| +{(c}_+*{sum}_+)*\left| S_G\right| ={(c}_+*{sum}_++c_**{sum}_*)*\left| S_G\right|\) , where \({sum}_*\) represents the number of multiplication gates in the circuit and \({sum}_+\) represents the number of multiplication gates in the circuit.

Computation cost

We assume \(T_{exp},T_{add},T_{mul},T_{hash},T_{mod},T_{pair}\) represent exponentiation operation, addition operation, multiplication operation, hash operation, module operation, and pairing operation of bilinear mapping respectively.

The calculating cost of Sign() is \(2T_{exp}\) .

In the GateVal(), the calculating cost is \(2T_{exp}\) for Eq. ( 4 ), the calculating cost is \({2T}_{add}+3T_{mul}+2T_{hash}+3T_{exp}\) for Eq. ( 5 ), the calculating cost is \(2T_{exp}+T_{add}+2T_{mul}+T_{hash}\) for Eq. ( 6 ), the calculating cost is \({2T}_{exp}\) for Eq. ( 7 ), the calculating cost is \({2T}_{mul}\) for Eq. ( 8 ).

The calculating cost of ProofGen() is the sum of the calculating costs of all gates \(\sum _{g\in \left| f\right| } T_g\) , where \(\left| f\right|\) represents the set of gates in a polynomial function, and \(T_g\) represents the calculating cost of performing an addition or multiplication gate of the GateVal() algorithm.

The calculating cost of VerifyProof() is \({2T}_{exp}+2T_{mul}+T_{pair}+T_\rho\) , where \(T_\rho\) indicates the calculating cost of computing \(\rho \leftarrow f\left( t_1,\ldots ,t_n\right)\) .

We compare the proposed scheme in the paper with existing solutions 11 , as shown in Table 3 . In Table 3 , we can see that their calculating costs are similar during the signature stage. However, in practice, the signature method of the two schemes are different in design and operation. There are two signature methods designed in the existing scheme 11 , which may be due to the requirement that the addition gate must be executed after the multiplication gate when generating the verification tag. This approach affects the generation efficiency of verification tag in subsequent steps. In contrast, the proposed scheme in the paper does not have this limitation, so it can generate verification tag more efficiently. In addition, the scheme in the paper uses Horner’s method to optimize polynomials when generating verification labels. We can see that the GateVal() and ProofCre() algorithms are used to generate verification tag, which is more efficient than existing scheme 11 . This method can make the generation of verification tag faster, thereby improving the efficiency of the entire scheme. In contrast, the existing schemes 11 may not adopt this optimization method, resulting in slower generation of verification tag. We can find that the difference in calculating cost between the two schemes mainly exists when the inputs are the variable \(x_1,x_2\) with the different public key. This is because for the polynomial-based verifiable computation scheme on multi-source outsourced data, the design of this part is the difficulty and focus. The scheme 11 require that addition gates must be executed after multiplication gates, and design complex two-level tags, resulting in inefficiency. The proposed scheme improves this by using a unified verification tag, allowing multiplication and addition gates to be executed in parallel, resulting in improved efficiency of verification tag generation.

To sum up, we use Horner’s method to optimize the polynomial, which will make the system execute the addition gate or multiplication gate significantly less times than the existing scheme. This is our main idea to improve efficiency. In order to adapt to this structure, an efficient verification tag is designed to support addition homomorphism and multiplication homomorphism, so that the addition gate and multiplication gate can be crossed and the verification speed can be further improved.As a result, the proposed scheme on efficiency is better than the existing scheme.

Experimental results

Experiment setup.

The experiment was conducted in the environment of Intel(R) Core(TM) i5-10210U CPU @ 1.60 GHz 2.11 GHz. The dataset we used was air pollution data for 367 major cities in China. In the experiment, we uploaded air pollution data as raw data to cloud server. This air pollution data will be used for data analysis, and the results of the data analysis will be published. These data analyses will be used to calculate the average air quality of all cities in the country, the average value of a certain pollution component in a city, etc.

The performance of the scheme is compared with homomorphic MAC 10 , ADSNARK 40 , and verifying tag 11 . The homomorphic MAC 10 does not support multiple data sources. We mainly measure the efficiency of the scheme from three aspects: the calculating cost of signature generation, the calculating cost of proof information generation, the calculating cost of verification phase, etc.

The calculating cost of signature generation

In the signature generation, the data is signed and uploaded to the cloud along with the original data. We must ensure the efficiency of signing, which is closely related to the efficiency of the offline phase. In order to verify the calculating cost of the signature generation algorithm Sign(), we compare the calculating cost of the Sign() algorithm in the scheme, the homomorphism MAC 10 , the ADSNARK 40 and verifying tag 11 , where the data scale ranges from 1000 to 10,000. As shown in the Fig. 3 , the calculating cost of signature increases with the increase of data volume. The time cost of signature generation in this scheme is similar to verifying tag 11 . Specially, the signature generation can be generated offline at the user side before the data is uploaded to the cloud, which will not affect the data correctness checks at the online stage. And Homomorphic MAC cannot directly support multiple data contributors, so this scheme can generate signatures relatively quickly and the input of polynomials support multiple data sources.

The calculating cost of the signature generation with different data scale.

The calculating cost of proof information generation

After the signature is generated, the cloud generates the proof information by algorithm ProofGen(). Here we evaluate the calculating cost of the algorithm ProofGen() from two perspectives.

First, one of the important motivations of our scheme is to improve efficiency and to be suitable for polynomial calculations of higher order. So we evaluate the calculating cost of polynomial functions with different order sizes. In order to comprehensively measure the efficiency of the scheme, we randomly select some polynomials with high order, where the order of polynomial function range from 50 to 500. We compare the ProofGen() algorithm in the scheme, the homomorphism MAC , the ADSNARK and verifying tag, and we assume that all data is outsourced and signed by a single user.

Figure 4 illustrates the calculating cost of the scheme will not increase obviously with the increase of polynomial order. And this scheme consumes less time compared with the other schemes 10 , 11 , 40 , which shows this scheme can effectively reduce the calculating cost of proof information generation. In particular, the scheme can generate verification tag much faster when the order of polynomials is high. This is because the scheme is not limited by the polynomial structure, and the multiplicative gate and the additive gates can be executed interactively. When polynomials are optimized by Horner’s Method, the verification tag generation can be faster compared to other schemes.

The calculating cost of the proof information generation with different polynomial order.

Second, one of the motivations for our scheme is that it is suitable for cases where the input of a polynomial comes from multiple data sources. So we evaluate the calculating cost of polynomial functions with different number of data owners, where the number of data owners ranges from 1 to 300. In the experiment, the input of the polynomial is pollution data from multiple cities, which will be signed using different private keys. The homomorphic MAC cannot directly support multiple data contributors, so we compare the ProofGen() algorithm in the scheme, the ADSNARK and verifying tag.

Figure 5 illustrates the algorithm ProofGen() is executed very quickly on the cloud server and does not significantly increase even the data sourcing from multiple users. It is very obvious that the scheme can perform calculations much faster than the other schemes. By comparing two subgraphs, the time cost does not increase significantly with the increase of polynomial order, which indicates that the scheme is suitable for complex polynomials. Therefore, this scheme can verify the correctness of the calculated results very quickly, in which the data can be derived from multiple data sources.

The calculating cost of proof information generation in the cloud with different number of data owner.

The calculating cost of verification phase

After the proof information is sent to the Key management center, the Key management center executes the algorithm VerifyProof() to verify the accuracy of the calculation results using the proof information. We evaluate the calculating cost of verification phase with different order sizes of polynomial function, and we assume that all data is outsourced and signed by a single user. We compare the ProofGen() algorithm in the scheme, the homomorphism MAC , the ADSNARK and verifying tag, and we assume that all data is outsourced and signed by a single user.

Figure 6 illustrates that the calculating cost of the scheme does not increase significantly with the increase of polynomial order. This is because even if the data comes from a large number of users, only one proof information corresponding to one result can be generated after polynomial calculation. Therefore, the time consumption during the verification phase does not significantly increase due to the increase in data sources. Then it turns out that the calculating cost of the scheme is smaller than that of other schemes. The experimental results show that the proposed scheme can quickly and effectively verify the correctness of the calculated results, even if the data comes from multiple data sources.

The calculating cost of the verification phase with different polynomial order.

This paper presents an efficient polynomial-based verifiable computation scheme for multi-source outsourced data. We optimize polynomials for faster verification using the Horner method, where addition and multiplication gates can interlace polynomial functions. In order to adapt to this structure, we design the corresponding homomorphic verification labels, so that the input of the polynomial can come from multiple data sources. Our proposal has some important advantages. First, it works with multi-source data, which means that the values of the input polynomials can come from multiple users. This can be important in practical applications, such as in distributed systems or secure multi-party computing. Secondly, our solution is efficient. By using the Horner method, we can reduce the amount of computation required, which speeds up verification.

However, our proposal also has some potential limitations and directions for future research. First, our scheme is only suitable for verifying the correctness of the calculation results of entrusted polynomial functions, and may not be suitable for all types of data and computation tasks. Then, the work of user in the preprocessing stage may be complex, and it will consume a certain amount of computing resources and storage resources in user side. To solve these problems, the future work will focus on extending this scheme to handle more complex polynomial functions and to further enhance its efficiency. We also plan to investigate the application of this scheme in other fields, such as cryptography and distributed computing, where polynomial-based computations play a crucial role. Additionally, we aim to develop more secure and privacy-preserving methods for outsourced data computation to address the concerns of untrustworthy cloud servers.

Verifiable computing means that the computing task is outsourced to the untrusted cloud server, and the untrusted cloud server needs to submit a correctness proof of the calculation results while completing the computing task. There are two main problems with the existing verifiable computing scheme. First, the existing scheme requires that the input of the polynomial must come from a single data source. Secondly, the design of verification labels may cause problems such as reduced efficiency, especially when the polynomial function is relatively complex, so that the verification process will be extremely slow, and even affect the use of data. To solve these problems, we design for the first time an efficient polynomial-based verifiable computation scheme on multi-source outsourced data, which has the characteristics of efficient and supporting multiple data sources. As the polynomial function is executed gate by gate, we use the key management center to convert the signatures signed by different user into the verification tag with the unified public and private keys, so that the input of the polynomial can come from multiple data sources. Specially, we optimize the polynomials using Horner’s Method, and the generation of corresponding verification tag can be generated with the cross-operator of multiplication gate and addition gate, so as to improve the efficiency. Then we demonstrate the security of the scheme from two aspects: correctness and soundness. The performance of the scheme is verified by experiments, which shows that the scheme is more efficient than the existing schemes. Therefore, the scheme is able to provide efficient verifiable computing services in cloud outsourcing services, where the input of polynomials can come from multiple data sources. Overall, the work presented in this paper represents a significant step forward in achieving efficient and secure verifiable computation on multi-source outsourced data. We believe that our future research will further enhance the capabilities and applicability of this scheme, paving the way for more reliable and privacy-preserving cloud computing services.

Data availability

The data used in the experiment can be downloaded at http://www.cnemc.cn/ . Moreover, more data used and analysed during the current study available from the corresponding author on reasonable request.

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Acknowledgements

This research was funded by Research and Verification of Key Technologies for Secure and Efficient Federated Learning grant number U22B2038.

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These authors contributed equally: Yiran Zhang and Huizheng Geng.

Authors and Affiliations

China Mobile Research Institute, Beijing, 100053, China

Yiran Zhang, Huizheng Geng, Li Su, Shen He & Li Lu

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Conceptualization, Y.Z. and H.G.; methodology, Y.Z. and H.G.; software, Y.Z.; validation, Y.Z. and H.G. and L.S.; formal analysis, S.H.; investigation, L.L.; resources, S.H.; data curation, Y.Z. and H.G.; writing—original draft preparation, Y.Z. and H.G.; writing—review and editing, Y.Z. and H.G.; visualization, Y.Z. and H.G.; supervision, Y.Z. and H.G.; project administration, Y.Z. and H.G.; funding acquisition, L.S. and S.H. All authors have read and agreed to the published version of the manuscript.

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Correspondence to Huizheng Geng .

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Zhang, Y., Geng, H., Su, L. et al. An efficient polynomial-based verifiable computation scheme on multi-source outsourced data. Sci Rep 14 , 8512 (2024). https://doi.org/10.1038/s41598-024-53267-x

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