problem solving on factoring polynomials

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4.4: Solve Polynomial Equations by Factoring

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• Page ID 6257

Learning Objectives

• Review general strategies for factoring.
• Solve polynomial equations by factoring.
• Find roots of a polynomial function.
• Find polynomial equations given the solutions.

Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials.

general guidelines for factoring polynomials

Step 1: Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Step 2: Determine the number of terms in the polynomial.

• Factor four-term polynomials by grouping.
• Factor trinomials (3 terms) using “trial and error” or the AC method.
• Difference of squares :\(a^{2}−b^{2}=(a+b)(a−b)\)
• Sum of squares : \(a^{2}+b^{2}\)  no general formula
• Difference of cubes : \(a^{3}−b^{3}=(a−b)(a^{2}+ab+b^{2})\)
• Sum of cubes : \(a^{3}+b^{3}=(a+b)(a^{2}−ab+b^{2})\)

Step 3: Look for factors that can be factored further.

Step 4: Check by multiplying.

If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization. In addition, not all polynomials with integer coefficients factor. When this is the case, we say that the polynomial is prime.

If an expression has a GCF, then factor this out first. Doing so is often overlooked and typically results in factors that are easier to work with. Furthermore, look for the resulting factors to factor further; many factoring problems require more than one step. A polynomial is completely factored when none of the factors can be factored further.

Example \(\PageIndex{1}\):

Factor \(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x\).

This four-term polynomial has a GCF of \(2x\). Factor this out first.

\(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x = 2 x \left( 27 x ^ { 3 } - 18 x ^ { 2 } - 12 x + 8 \right)\)

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

\(2 x ( 3 x - 2 ) ^ { 2 } ( 3 x + 2 )\). The check is left to the reader.

Example \(\PageIndex{2}\):

Factor: \(x ^ { 4 } - 3 x ^ { 2 } - 4\).

This trinomial does not have a GCF.

\(\begin{aligned} x ^ { 4 } - 3 x ^ { 2 } - 4 & = \left( x ^ { 2 } \quad\right) \left( x ^ { 2 }\quad \right) \\ & = \left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } - 4 \right) \quad\color{Cerulean} { Difference\: of\: squares } \\ & = \left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 ) \end{aligned}\)

The factor \(\left( x ^ { 2 } + 1 \right)\) is prime and the trinomial is completlely factored.

\(\left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 )\)

Example \(\PageIndex{3}\):

Factor: \(x ^ { 6 } + 6 x ^ { 3 } - 16\).

Begin by factoring \(x ^ { 6 } = x ^ { 3 } \cdot x ^ { 3 }\) and look for the factors of \(16\) that add to \(6\).

\(\begin{aligned} x ^ { 6 } + 6 x ^ { 3 } - 16 & = \left( x ^ { 3 }\quad \right) \left( x ^ { 3 }\quad \right) \\ & = \left( x ^ { 3 } - 2 \right) \left( x ^ { 3 } + 8 \right)\quad \color{Cerulean} { sum\:of\:cubes } \\ & = \left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } - 2 x + 4 \right) \end{aligned}\)

The factor \(\left( x ^ { 3 } - 2 \right)\) cannot be factored any further using integers and the factorization is complete.

\(\left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

Exercise \(\PageIndex{1}\)

Factor: \(9 x ^ { 4 } + 17 x ^ { 2 } - 2\)

\(( 3 x + 1 ) ( 3 x - 1 ) \left( x ^ { 2 } + 2 \right)\)

www.youtube.com/v/lsAP8_UgUx0

Solving Polynomial Equations by Factoring

In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property 20 :

\(a⋅b=0\)   if and only if   \(a=0\) or \(b=0\)

The zero-product property is true for any number of factors that make up an equation. In other words, if any product is equal to zero, then at least one of the variable factors must be equal to zero. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Example \(\PageIndex{4}\)

Solve: \(2 x ( x - 4 ) ( 5 x + 3 ) = 0\).

Set each variable factor equal to zero and solve.

\(\begin{aligned} 2 x & = 0 \\ \frac { 2 x } { \color{Cerulean}{2} } & = \frac { 0 } {\color{Cerulean}{ 2} } \\ x & = 0 \end{aligned}\) or \(\begin{aligned} x - 4 & = 0 \\ x & = 4 \end{aligned}\) or \(\begin{aligned} 5 x + 3 & = 0 \\ \frac { 5 x } { \color{Cerulean}{5} } & = \frac { - 3 } {\color{Cerulean}{ 5} } \\ x & = - \frac { 3 } { 5 } \end{aligned}\)

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

The solutions are \(0, 4\), and \(\frac{−3}{5}\).

Of course, most equations will not be given in factored form.

Example \(\PageIndex{5}\)

Solve: \(4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 = 0\).

Begin by factoring the left side completely.

\(\begin{aligned} 4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 & = 0 \quad\color{Cerulean}{Factor\:by\:grouping.} \\ x ^ { 2 } ( 4 x - 1 ) - 25 ( 4 x - 1 ) & = 0 \\ ( 4 x - 1 ) \left( x ^ { 2 } - 25 \right) & = 0\quad\color{Cerulean}{Factor\:as\:a\:difference\:of\:squares.} \\ ( 4 x - 1 ) ( x + 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Set each factor equal to zero and solve.

\(\begin{array} { r } { 4 x - 1 = 0 } \\ { 4 x = 1 } \\ { x = \frac { 1 } { 4 } } \end{array}\) or \(\begin{aligned} x + 5 & = 0 \\ x & = - 5 \end{aligned}\) or \(\begin{aligned} x - 5 & = 0 \\ x & = 5 \end{aligned}\)

The solutions are \(\frac{1}{4}, −5\), and \(5\).

Using the zero-product property after factoring an equation that is equal to zero is the key to this technique. However, the equation may not be given equal to zero, and so there may be some preliminary steps before factoring. The steps required to solve by factoring 21 are outlined in the following example.

Example \(\PageIndex{6}\)

Solve: \(15 x ^ { 2 } + 3 x - 8 = 5 x - 7\).

Step 1: Express the equation in standard form, equal to zero. In this example, subtract \(5x\) from and add \(7\) to both sides.

\(\begin{array} { l } { 15 x ^ { 2 } + 3 x - 8 = 5 x - 7 } \\ { 15 x ^ { 2 } - 2 x - 1 = 0 } \end{array}\)

Step 2: Factor the expression.

\((3x−1)(5x+1)=0\)

Step 3: Apply the zero-product property and set each variable factor equal to zero.

\(3x−1=0\)        or        \(5x+1=0\)

Step 4: Solve the resulting linear equations.

\(\begin{array} { r } { 3 x - 1 = 0 } \\ { 3 x = 1 } \\ { x = \frac { 1 } { 3 } } \end{array}\) or \(\begin{aligned} 5 x + 1 & = 0 \\ 5 x & = - 1 \\ x & = - \frac { 1 } { 5 } \end{aligned}\)

The solutions are \(\frac{1}{3}\) and \(\frac{−1}{5}\). The check is optional.

Example \(\PageIndex{7}\):

Solve: \((3x+2)(x+1)=4\).

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to \(4\). However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

\(\begin{array} { r } { ( 3 x + 2 ) ( x + 1 ) = 4 } \\ { 3 x ^ { 2 } + 3 x + 2 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x - 2 = 0 } \end{array}\)

Once it is in standard form, we can factor and then set each factor equal to zero.

        \(\begin{array} { c } { ( 3 x - 1 ) ( x + 2 ) = 0 } \\ { 3 x - 1 = 0 \quad \text { or } \quad x + 2 = 0 } \\ \quad\quad{ 3 x = 1 }\quad\quad\quad\:\:\quad x=-2\\ { x = \frac { 1 } { 3 } }\quad\quad\quad\quad\quad\: \end{array}\)

The solutions are \(\frac{1}{3}\) and \(−2\).

Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

\(f ( x ) = a _ { n } x ^ { n } + a _ { n - 1 } x ^ { n - 1 } + \cdots + a _ { 1 } x + a _ { 0 }\)

A root 22 of a function is a value in the domain that results in zero. In other words, the roots occur when the function is equal to zero, \(f(x)=0\).

Example \(\PageIndex{8}\)

Find the roots: \(f ( x ) = ( x + 2 ) ^ { 2 } - 4\).

To find roots we set the function equal to zero and solve.

\(\begin{aligned} f ( x ) & = 0 \\ ( x + 2 ) ^ { 2 } - 4 & = 0 \\ x ^ { 2 } + 4 x + 4 - 4 & = 0 \\ x ^ { 2 } + 4 x & = 0 \\ x ( x + 4 ) & = 0 \end{aligned}\)

Next, set each factor equal to zero and solve.

\(\begin{aligned} x = 0 \quad \text { or } \quad x + 4 = 0 \\ x = - 4 \end{aligned}\)

We can show that these \(x\)-values are roots by evaluating.

\(\begin{aligned} f ( 0 ) & = ( 0 + 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)   \(\begin{aligned} f ( - 4 ) & = ( - 4 + 2 ) ^ { 2 } - 4 \\ & = ( - 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)

The roots are \(0\) and \(−4\).

If we graph the function in the previous example we will see that the roots correspond to the \(x\)-intercepts of the function. Here the function \(f\) is a basic parabola shifted \(2\) units to the left and \(4\) units down.

Example \(\PageIndex{9}\)

Find the roots: \(f ( x ) = x ^ { 4 } - 5 x ^ { 2 } + 4\).

\(\begin{aligned} f ( x ) & = 0 \\ x ^ { 4 } - 5 x ^ { 2 } + 4 & = 0 \\ \left( x ^ { 2 } - 1 \right) \left( x ^ { 2 } - 4 \right) & = 0 \\ ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \end{aligned}\)

\(\begin{aligned} x + 1 & = 0 \\ x & = - 1 \end{aligned}\) or \(\begin{array} { r } { x - 1 = 0 } \\ { x = 1 } \end{array}\) or \(\begin{aligned} x + 2 & = 0 \\ x & = - 2 \end{aligned}\) or \(\begin{aligned} x - 2 & = 0 \\ x & = 2 \end{aligned}\)

The roots are \(−1, 1, −2\), and \(2\).

Graphing the previous function is not within the scope of this course. However, the graph is provided below:

Notice that the degree of the polynomial is \(4\) and we obtained four roots. In general, for any polynomial function with one variable of degree \(n\), the fundamental theorem of algebra 23 guarantees \(n\) real roots or fewer. We have seen that many polynomials do not factor. This does not imply that functions involving these unfactorable polynomials do not have real roots. In fact, many polynomial functions that do not factor do have real solutions. We will learn how to find these types of roots as we continue in our study of algebra.

Example \(\PageIndex{10}\)

Find the roots: \(f ( x ) = - x ^ { 2 } + 10 x - 25\).

\(\begin{aligned} f ( x ) & = 0 \\ - x ^ { 2 } + 10 x - 25 & = 0 \\ - \left( x ^ { 2 } - 10 x + 25 \right) & = 0 \\ - ( x - 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Next, set each variable factor equal to zero and solve.

\(\begin{aligned} x - 5 & = 0 \quad\quad\text { or }& x - 5 = 0 \\ & = 5 \quad &x= 5 \end{aligned}\)

A solution that is repeated twice is called a double root 24 . In this case, there is only one solution.

The root is \(5\).

The previous example shows that a function of degree \(2\) can have one root. From the factoring step, we see that the function can be written

\(f ( x ) = - ( x - 5 ) ^ { 2 }\)

In this form, we can see a reflection about the \(x\)-axis and a shift to the right \(5\) units. The vertex is the \(x\)-intercept, illustrating the fact that there is only one root.

Exercise \(\PageIndex{2}\)

Find the roots of \(f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - x - 3\).

\(±1, −3\)

www.youtube.com/v/t1ShqqhoaVE

Example \(\PageIndex{11}\)

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\), where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(40\) feet.

We are asked to find the speed \(x\) where the safe stopping distance \(d(x)=40\) feet.

\(\begin{array} { c } { d ( x ) = 40 } \\ { \frac { 1 } { 20 } x ^ { 2 } + x = 40 } \end{array}\)

To solve for \(x\), rewrite the resulting equation in standard form. In this case, we will first multiply both sides by \(20\) to clear the fraction.

\(\begin{aligned} \color{Cerulean}{20}\color{black}{ \left( \frac { 1 } { 20 } x ^ { 2 } + x \right)} & = \color{Cerulean}{20}\color{black}{ (} 40 ) \\ x ^ { 2 } + 20 x & = 800 \\ x ^ { 2 } + 20 x - 800 & = 0 \end{aligned}\)

Next factor and then set each factor equal to zero.

\(\begin{aligned} x ^ { 2 } + 20 x - 800 & = 0 \\ ( x + 40 ) ( x - 20 ) & = 0 \\ x + 40 & = 0\quad or\quad \:x-20=0 \\ x & = - 40\quad \quad\quad\quad x=20 \end{aligned}\)

The negative answer does not make sense in the context of this problem. Consider \(x=20\) miles per hour to be the only solution.

\(20\) miles per hour

Finding Equations with Given Solutions

We can use the zero-product property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Example \(\PageIndex{12}\)

Find a quadratic equation with integer coefficients, given solutions \(\frac{−3}{2}\) and \(\frac{1}{3}\).

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

\(\begin{aligned} x & = - \frac { 3 } { 2 } \\ 2 x & = - 3 \\ 2 x + 3 & = 0 \end{aligned}\) or \(\begin{aligned} x & = \frac { 1 } { 3 } \\ 3 x & = 1 \\ 3 x - 1 & = 0 \end{aligned}\)

The product of these linear factors is equal to zero when \(x=\frac{−3}{2}\) or \(x=\frac{1}{3}\).

\((2x+3)(3x−1)=0\)

Multiply the binomials and present the equation in standard form.

\(\begin{array} { r } { 6 x ^ { 2 } - 2 x + 9 x - 3 = 0 } \\ { 6 x ^ { 2 } + 7 x - 3 = 0 } \end{array}\)

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

\(6 x ^ { 2 } + 7 x - 3 = 0\)

Example \(\PageIndex{13}\)

Find a polynomial function with real roots \(1, −2\), and \(2\).

Given solutions to \(f(x)=0\) we can find linear factors.

\(\begin{array} { r } { x = 1 } \\ { x - 1 = 0 } \end{array}\) or \(\begin{aligned} x & = - 2 \\ x + 2 & = 0 \end{aligned}\) or \(\begin{aligned} x & = 2 \\ x - 2 & = 0 \end{aligned}\)

Apply the zero-product property and multiply.

\(\begin{aligned} ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \\ ( x - 1 ) \left( x ^ { 2 } - 4 \right) & = 0 \\ x ^ { 3 } - 4 x - x ^ { 2 } + 4 & = 0 \\ x ^ { 3 } - x ^ { 2 } - 4 x + 4 & = 0 \end{aligned}\)

\(f ( x ) = x ^ { 3 } - x ^ { 2 } - 4 x + 4\)

Exercise \(\PageIndex{3}\)

Find a polynomial equation with integer coefficients, given solutions \(\frac{1}{2}\) and \(\frac{−3}{4}\).

\(8 x ^ { 2 } + 2 x - 3 = 0\)

www.youtube.com/v/o4cuUWWEGdU

Key Takeaways

• Factoring and the zero-product property allow us to solve equations.
• To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting equations are the solutions to the original.
• Not all polynomial equations can be solved by factoring. We will learn how to solve polynomial equations that do not factor later in the course.
• A polynomial function can have at most a number of real roots equal to its degree. To find roots of a function, set it equal to zero and solve.
• To find a polynomial equation with given solutions, perform the process of solving by factoring in reverse.

Exercise \(\PageIndex{4}\)

Factor completely.

• \(50 x ^ { 2 } - 18\)
• \(12 x ^ { 3 } - 3 x\)
• \(10 x ^ { 3 } + 65 x ^ { 2 } - 35 x\)
• \(15 x ^ { 4 } + 7 x ^ { 3 } - 4 x ^ { 2 }\)
• \(6 a ^ { 4 } b - 15 a ^ { 3 } b ^ { 2 } - 9 a ^ { 2 } b ^ { 3 }\)
• \(8 a ^ { 3 } b - 44 a ^ { 2 } b ^ { 2 } + 20 a b ^ { 3 }\)
• \(36 x ^ { 4 } - 72 x ^ { 3 } - 4 x ^ { 2 } + 8 x\)
• \(20 x ^ { 4 } + 60 x ^ { 3 } - 5 x ^ { 2 } - 15 x\)
• \(3 x ^ { 5 } + 2 x ^ { 4 } - 12 x ^ { 3 } - 8 x ^ { 2 }\)
• \(10 x ^ { 5 } - 4 x ^ { 4 } - 90 x ^ { 3 } + 36 x ^ { 2 }\)
• \(x ^ { 4 } - 23 x ^ { 2 } - 50\)
• \(2 x ^ { 4 } - 31 x ^ { 2 } - 16\)
• \(- 2 x ^ { 5 } - 6 x ^ { 3 } + 8 x\)
• \(- 36 x ^ { 5 } + 69 x ^ { 3 } + 27 x\)
• \(54 x ^ { 5 } - 78 x ^ { 3 } + 24 x\)
• \(4 x ^ { 6 } - 65 x ^ { 4 } + 16 x ^ { 2 }\)
• \(x ^ { 6 } - 7 x ^ { 3 } - 8\)
• \(x ^ { 6 } - 25 x ^ { 3 } - 54\)
• \(3 x ^ { 6 } + 4 x ^ { 3 } + 1\)
• \(27 x ^ { 6 } - 28 x ^ { 3 } + 1\)

1. \(2 ( 5 x + 3 ) ( 5 x - 3 )\)

3. \(5 x ( x + 7 ) ( 2 x - 1 )\)

5. \(3 a ^ { 2 } b ( 2 a + b ) ( a - 3 b )\)

7. \(4 x ( x - 2 ) ( 3 x + 1 ) ( 3 x - 1 )\)

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

11. \(\left( x ^ { 2 } + 2 \right) ( x + 5 ) ( x - 5 )\)

13. \(- 2 x \left( x ^ { 2 } + 4 \right) ( x - 1 ) ( x + 1 )\)

15. \(6 x ( x + 1 ) ( x - 1 ) ( 3 x + 2 ) ( 3 x - 2 )\)

17. \(( x + 1 ) \left( x ^ { 2 } - x + 1 \right) ( x - 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

19. \(\left( 3 x ^ { 3 } + 1 \right) ( x + 1 ) \left( x ^ { 2 } - x + 1 \right)\)

Exercise \(\PageIndex{5}\)

• \(( 6 x - 5 ) ( x + 7 ) = 0\)
• \(( x + 9 ) ( 3 x - 8 ) = 0\)
• \(5 x ( 2 x - 5 ) ( 3 x + 1 ) = 0\)
• \(4 x ( 5 x - 1 ) ( 2 x + 3 ) = 0\)
• \(( x - 1 ) ( 2 x + 1 ) ( 3 x - 5 ) = 0\)
• \(( x + 6 ) ( 5 x - 2 ) ( 2 x + 9 ) = 0\)
• \(( x + 4 ) ( x - 2 ) = 16\)
• \(( x + 1 ) ( x - 7 ) = 9\)
• \(( 6 x + 1 ) ( x + 1 ) = 6\)
• \(( 2 x - 1 ) ( x - 4 ) = 39\)
• \(x ^ { 2 } - 15 x + 50 = 0\)
• \(x ^ { 2 } + 10 x - 24 = 0\)
• \(3 x ^ { 2 } + 2 x - 5 = 0\)
• \(2 x ^ { 2 } + 9 x + 7 = 0\)
• \(\frac { 1 } { 10 } x ^ { 2 } - \frac { 7 } { 15 } x - \frac { 1 } { 6 } = 0\)
• \(\frac { 1 } { 4 } - \frac { 4 } { 9 } x ^ { 2 } = 0\)
• \(6 x ^ { 2 } - 5 x - 2 = 30 x + 4\)
• \(6 x ^ { 2 } - 9 x + 15 = 20 x - 13\)
• \(5 x ^ { 2 } - 23 x + 12 = 4 ( 5 x - 3 )\)
• \(4 x ^ { 2 } + 5 x - 5 = 15 ( 3 - 2 x )\)
• \(( x + 6 ) ( x - 10 ) = 4 ( x - 18 )\)
• \(( x + 4 ) ( x - 6 ) = 2 ( x + 4 )\)
• \(4 x ^ { 3 } - 14 x ^ { 2 } - 30 x = 0\)
• \(9 x ^ { 3 } + 48 x ^ { 2 } - 36 x = 0\)
• \(\frac { 1 } { 3 } x ^ { 3 } - \frac { 3 } { 4 } x = 0\)
• \(\frac { 1 } { 2 } x ^ { 3 } - \frac { 1 } { 50 } x = 0\)
• \(- 10 x ^ { 3 } - 28 x ^ { 2 } + 48 x = 0\)
• \(- 2 x ^ { 3 } + 15 x ^ { 2 } + 50 x = 0\)
• \(2 x ^ { 3 } - x ^ { 2 } - 72 x + 36 = 0\)
• \(4 x ^ { 3 } - 32 x ^ { 2 } - 9 x + 72 = 0\)
• \(45 x ^ { 3 } - 9 x ^ { 2 } - 5 x + 1 = 0\)
• \(x ^ { 3 } - 3 x ^ { 2 } - x + 3 = 0\)
• \(x ^ { 4 } - 5 x ^ { 2 } + 4 = 0\)
• \(4 x ^ { 4 } - 37 x ^ { 2 } + 9 = 0\)

1. \(- 7 , \frac { 5 } { 6 }\)

3. \(0 , \frac { 5 } { 2 } , - \frac { 1 } { 3 }\)

5. \(- \frac { 1 } { 2 } , 1 , \frac { 5 } { 3 }\)

7. \(- 6,4\)

9. \(- \frac { 5 } { 3 } , \frac { 1 } { 2 }\)

11. \(5,10\)

13. \(- \frac { 5 } { 3 } , 1\)

15. \(- \frac { 1 } { 3 } , 5\)

17. \(- \frac { 1 } { 6 } , 6\)

19. \(\frac { 3 } { 5 } , 8\)

21. \(2,6\)

23. \(0 , - \frac { 3 } { 2 } , 5\)

25. \(0 , \pm \frac { 3 } { 2 }\)

27. \(- 4,0 , \frac { 6 } { 5 }\)

29. \(\pm 6 , \frac { 1 } { 2 }\)

31. \(\pm \frac { 1 } { 3 } , \frac { 1 } { 5 }\)

33. \(\pm 1 , \pm 2\)

Exercise \(\PageIndex{6}\)

Find the roots of the given functions.

• \(f ( x ) = x ^ { 2 } + 10 x - 24\)
• \(f ( x ) = x ^ { 2 } - 14 x + 48\)
• \(f ( x ) = - 2 x ^ { 2 } + 7 x + 4\)
• \(f ( x ) = - 3 x ^ { 2 } + 14 x + 5\)
• \(f ( x ) = 16 x ^ { 2 } - 40 x + 25\)
• \(f ( x ) = 9 x ^ { 2 } - 12 x + 4\)
• \(g ( x ) = 8 x ^ { 2 } + 3 x\)
• \(g ( x ) = 5 x ^ { 2 } - 30 x\)
• \(p ( x ) = 64 x ^ { 2 } - 1\)
• \(q ( x ) = 4 x ^ { 2 } - 121\)
• \(f ( x ) = \frac { 1 } { 5 } x ^ { 3 } - 1 x ^ { 2 } - \frac { 1 } { 20 } x + \frac { 1 } { 4 }\)
• \(f ( x ) = \frac { 1 } { 3 } x ^ { 3 } + \frac { 1 } { 2 } x ^ { 2 } - \frac { 4 } { 3 } x - 2\)
• \(g ( x ) = x ^ { 4 } - 13 x ^ { 2 } + 36\)
• \(g ( x ) = 4 x ^ { 4 } - 13 x ^ { 2 } + 9\)
• \(f ( x ) = ( x + 5 ) ^ { 2 } - 1\)
• \(g ( x ) = - ( x + 5 ) ^ { 2 } + 9\)
• \(f ( x ) = - ( 3 x - 5 ) ^ { 2 }\)
• \(g ( x ) = - ( x + 2 ) ^ { 2 } + 4\)

1. \(2 , - 12\)

3. \(- \frac { 1 } { 2 } , 4\)

5. \(\frac { 5 } { 4 }\)

7. \(- \frac { 3 } { 8 } , 0\)

9. \(\pm \frac { 1 } { 8 }\)

11. \(\pm \frac { 1 } { 2 } , 5\)

13. \(\pm 2 , \pm 3\)

15. \(- 6 , - 4\)

17. \(\frac { 5 } { 3 }\)

Exercise \(\PageIndex{7}\)

Given the graph of a function, determine the real roots.

5. The sides of a square measure \(x − 2\) units. If the area is \(36\) square units, then find \(x\).

6. The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

7. The profit in dollars generated by producing and selling n bicycles per week is given by the formula \(P ( n ) = - 5 n ^ { 2 } + 400 n - 6000\). How many bicycles must be produced and sold to break even?

8. The height in feet of an object dropped from the top of a \(64\)-foot building is given by \(h ( t ) = - 16 t ^ { 2 } + 64\) where \(t\) represents the time in seconds after it is dropped. How long will it take to hit the ground?

9. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

What is the length of each side of the cardboard sheet if the volume of the box is to be \(98\) cubic inches?

10. The height of a triangle is \(4\) centimeters less than twice the length of its base. If the total area of the triangle is \(48\) square centimeters, then find the lengths of the base and height.

11. A uniform border is to be placed around an \(8 × 10\) inch picture.

If the total area including the border must be \(168\) square inches, then how wide should the border be?

12. The area of a picture frame including a \(3\)-inch wide border is \(120\) square inches.

If the width of the inner area is \(2\) inches less than its length, then find the dimensions of the inner area.

13. Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\) where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(75\) feet.

14. A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula \(R ( n ) = 12 n - 0.6 n ^ { 2 }\) where \(n\) represents the number of palettes of product sold \((0 ≤ n < 20)\). Determine the number of palettes sold in a day if the revenue was \(45\) thousand dollars.

1. \(- 3 , - 1,0,2\)

3. \(- 2,3\)

5. \(8\) units

7. \(20\) or \(60\) bicycles

9. \(11\) in

11. \(2\) inches

13. \(30\) miles per hour

Exercise \(\PageIndex{8}\)

Find a polynomial equation with the given solutions.

• \(2, \frac{1}{3}\)
• \(- \frac { 3 } { 4 } , 5\)
• \(- 3,1,3\)
• \(- 5 , - 1,1\)

1. \(x ^ { 2 } - 2 x - 15 = 0\)

3. \(3 x ^ { 2 } - 7 x + 2 = 0\)

5. \(x ^ { 2 } + 4 x = 0\)

7. \(x ^ { 2 } - 49 = 0\)

9. \(x ^ { 3 } - x ^ { 2 } - 9 x + 9 = 0\)

Exercise \(\PageIndex{9}\)

Find a function with the given roots.

• \(\frac { 1 } { 2 } , \frac { 2 } { 3 }\)
• \(\frac { 2 } { 5 } , - \frac { 1 } { 3 }\)
• \(\pm \frac { 3 } { 4 }\)
• \(\pm \frac { 5 } { 2 }\)
• \(5\) double root
• \(-3\) double root

1. \(f ( x ) = 6 x ^ { 2 } - 7 x + 2\)

3. \(f ( x ) = 16 x ^ { 2 } - 9\)

5. \(f ( x ) = x ^ { 2 } - 10 x + 25\)

7. \(f ( x ) = x ^ { 3 } - 2 x ^ { 2 } - 3 x\)

Exercise \(\PageIndex{10}\)

Recall that if \(| X | = p\), then \(X=-p\) or \(X=p\). Use this to solve the following absolute value equations.

• \(\left| x ^ { 2 } - 8 \right| = 8\)
• \(\left| 2 x ^ { 2 } - 9 \right| = 9\)
• \(\left| x ^ { 2 } - 2 x - 1 \right| = 2\)
• \(\left| x ^ { 2 } - 8 x + 14 \right| = 2\)
• \(\left| 2 x ^ { 2 } - 4 x - 7 \right| = 9\)
• \(\left| x ^ { 2 } - 3 x - 9 \right| = 9\)

1. \(\pm 4,0\)

3. \(\pm 1,3\)

5. \(- 2,1,4\)

Exercise \(\PageIndex{11}\)

• Explain to a beginning algebra student the difference between an equation and an expression.
• What is the difference between a root and an \(x\)-intercept? Explain.
• Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.
• Research and discuss the fundamental theorem of algebra.

1. Answer may vary

3. Answer may vary

20 A product is equal to zero if and only if at least one of the factors is zero.

21 The process of solving an equation that is equal to zero by factoring it and then setting each variable factor equal to zero.

22 A value in the domain of a function that results in zero.

23 Guarantees that there will be as many (or fewer) roots to a polynomial function with one variable as its degree.

24 A root that is repeated twice.

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Factoring Polynomials

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• Ashlin Farris
• Redwan Mohammed

Factoring a polynomial is the process of decomposing a polynomial into a product of two or more polynomials. For example, \( f(x) = x^2 + 5x + 6 \) can be decomposed into \( f(x) = (x+3)(x+2) .\)

Another example:

Factor \(x^2 - x - 6 \). We have \[ x^2 - x - 6 = (x-3)(x+2).\ _\square\] Note that \( x^2 - x - 6 \) can also be expressed as \( 1 \cdot (x^2 -x - 6) \). Thus, the factors of \( x^2 - x - 6 \) are \(1\), \(x^2 - x - 6\), \(x-3\) and \(x+2\).

In order to solve problems like this, you will need to understand prime factorization . Review this concept, if needed, before continuing. We will look at 3 common ways in which a polynomial can be factored: grouping, substitution, and using identities.

Factoring Polynomials by Grouping

Factoring by substitution, factoring polynomials using identities.

We often see the grouping method applied to polynomials with 4 terms. The idea is to pair like terms together so that we can apply the distributive property in order to factorize them nicely.

Factor \( x^3 - 3x^2 -x + 3 \). We have \[ \begin{align} x^3 - 3x^2 - x + 3 &= x^2(x - 3) - (x - 3) \\ &= (x^2 -1)(x - 3) \\ &= (x - 1)(x + 1)(x - 3). \ _\square \end{align} \]

Factor \( x^2 + 4x - y^2 - 4y \). We have \[ \begin{align} x^2 + 4x - y^2 - 4y &= (x^2 - y^2) + (4x - 4y) \\ &= (x -y)(x + y) + 4(x-y) \\ &= (x - y)(x + y + 4). \ _\square \end{align} \]

Factor \( x^2 - 16y^2 - 6x + 9 \). We have \[ \begin{align} x^2 - 16y^2 - 6x + 9 &= (x^2 - 6x + 9) - 16y^2 \\ &= (x-3)^2 - (4y)^2 \\ &= (x - 3 - 4y)(x - 3 + 4y). \ _\square \end{align} \]

If a polynomial is complicated, you can try substituting one of the complicated terms with a simpler term to make it easier to factor.

Factor \( (x - y)(x - y - 1) - 20 \). Let \( S \) = \( x - y \). Substituting \( S \) in for \( x - y\) for ease of computation, we get \[ \begin{align} (x - y)(x - y - 1) - 20 &= (S)(S - 1) - 20 \\ &= S^2 - S - 20 \\ &= (S - 5)(S + 4)\\ &= (x - y - 5)(x - y + 4). \ _\square \end{align} \]

Factor \( (x + y)(x + y + 1) - 12 \). Let \( S \) = \( x + y \). Substituting \( S \) in for \( x + y\) for ease of computation, we get \[ \begin{align} (x + y)(x + y + 1) - 12 &= (S)(S + 1) - 12 \\ &= S^2 + S - 12\\ &= (S - 3)(S + 4)\\ &= (x + y - 3)(x + y + 4). \ _\square \end{align} \]

You can see more examples applying this method on the Factoring Polynomials by Substitution page.

Factoring binomials of the form \( x^2 - y^2= (x-y)(x+y):\) This approach applies the difference of two squares identity .

Factor \( x^2 - 16 \). We have \[ x^2 - 16 = x^2 - 4^2 = (x-4)(x+4).\ _\square\]

Factor \( x^4 - 25y^4 \). We have \[ x^4 - 25y^4 = \big(x^2\big)^2 - \big(5y^2\big)^2 = \big(x^2 - 5y^2\big)\big(x^2 + 5y^2\big).\ _\square\]

Factor \( 9x^3y - 36xy \). We have \[ 9x^3y - 36xy = 9xy\big(x^2 - 4\big) = 9xy\big(x^2 - 2^2\big) = 9xy(x-2)(x+2).\ _\square\]

Factor \( (x+1)^2 - 9(x-2)^2 \). We have \[ \begin{align} (x+1)^2 - 9(x-2)^2 &= (x+1)^2 - \big(3(x-2)\big)^2 \\ &= \big((x+1) - 3(x-2)\big) \big((x+1) + 3(x-2)\big) \\ &= (x + 1 -3x + 6)(x + 1 + 3x - 6) \\ &= (-2x + 7)(4x - 5). \ _\square \end{align} \]

Factoring binomials of the form \( x^3 + y^3 = (x+y)\big(x^2 - xy + y^2\big) \) and \(x^3 - y^3 = (x-y)\big(x^2 + xy+ y^2\big): \) This approach applies the sum and difference of cubes identity .

Factor \( 27x^3 - y^3 \). We have \[ \begin{align} 27x^3 - y^3 &= (3x)^3 - y^3 \\ &= (3x - y)\big((3x)^2 + 3xy + y^2\big) \\ &= (3x-y)(9x^2 + 3xy + y^2). \ _\square \end{align} \]

Factor \( 64x^3 + 27y^3 \). We have \[ \begin{align} 64x^3 + 27y^3 &= (4x)^3 + (3y)^3 \\ &= (4x + 3y)\big((4x)^2 - 12xy + (3y)^2\big) \\ &= (4x + 3y)\big(16x^2 -12 xy + 9y^2\big). \ _\square \end{align} \]

\[ \frac {2.3^3 - 0.027}{2.3^2 + 0.69 + 0.09} \]

Quick! Calculate the expression above as fast as possible! Time is of the essence!

Factoring trinomials of the form \( x^2 + 2xy + y^2 = (x+y)^2 \) and \(x^2 - 2xy + y^2 = (x-y)^2: \)

Factor \( x^2 + 6x + 9 \). We have \[ x^2 + 6x + 3 = x^2 + 2(x \cdot 3) + 3^2 = (x+3)^2. \ _\square \]

Factor \( x^2 - 4x + 4 \). We have \[ x^2 - 4x + 4 = x^2 - 2(x \cdot 2) + 2^2 = (x-2)^2. \ _\square \]

Factor \( 25x^2 - 20x + 4 \). We have \[ 25x^2 - 20x + 4 = (5x)^2 - 2(5x \cdot 2) + 2^2 = (5x-2)^2. \ _\square \]

Factor \( 16x^2 - 24xy + 9y^2 \). We have \[ 16x^2 - 24xy + 9y^2 = (4x)^2 - 2(4x \cdot 3y) + (3y)^2 = (4x-3y)^2. \ _\square \]

Factoring trinomials of the form \( x^2 + (a+b)x + ab = (x+a)(x+b): \) This approach is also known as factorization by observation.

In cases where we don't have a perfect square of the form \( (x+y)^2 \) or \( (x-y)^2 \), but the leading coefficient of \( x \) is 1, we can try to find \(a \) and \( b\) such that \( a \cdot b\) is equal to the constant term, and \( a + b \) is equal to the coefficient of \( x \).

Factor \( x^2 + 5x + 6 \). We have \[ x^2 + 5x + 6 = x^2 + (2 + 3)x + (2 \times 3) = (x + 2)(x + 3). \ _\square \]

Factor \( x^2 - 3x - 4 \). We have \[ x^2 - 3x + 4 = x^2 + (1 - 4)x + \big(1 \times (-4)\big) = (x + 1)(x - 4). \ _\square \]

Factor \( x^2 - 12x + 35 \). We have \[ x^2 - 12x + 35 = x^2 + (-5 - 7)x + \big((-5) \times (-7)\big) = (x - 5)(x - 7). \ _\square \]

Factor \( 2x^2 + 4xy - 30y^2 \). We have \[ \begin{align} 2x^2 + 4xy - 30y^2 &= 2\big(x^2 + 2xy - 15y^2\big) \\ &= 2\big(x^2 + (5y - 3y)x + (5y) \times (-3y)\big) \\ &= 2(x + 5y)(x - 3y) . \ _\square \end{align} \]

Factoring trinomials of the form \( acx^2 + (ad+bc)x + bd = (ax+b)(cx+d): \) When the leading coefficient of \(x\) is not 1, we must factor both the leading coefficient and the constant.

Factor \( 2x^2 + 4x + 2 \). Factors of \( 2x^2: \) \( 2x \) and \( x \) Factors of \( 2 :\) \( 1 \) and \( 2 \) Since we're looking at all-positive coefficients, we can ignore negative factors. Given this, two simple possible factorizations are \( (2x + 1)(x + 2) \) and \( (2x + 2)(x + 1) \) . In both of these factorizations, the leading term will be \(2x \cdot x = 2x^2 \) and the constant term will be \( 1 \cdot 2 = 2 \), So we don't have to worry about them. Let's turn our attention to getting the correct middle term \( 4x \). In the factorization \( (2x + 1)(x + 2) \), the middle term would be \( (2x \cdot 2) + (1 \cdot x) = 5x \), which is incorrect. In the factorization \( (2x + 2)(x + 1) \), the middle term would be \( (2x \cdot 1) + (2 \cdot x) = 4x \), which is what we want. So the factorization that gives us the correct expansion is \( (2x + 2)(x + 1) \). \(_\square\)

Factor \( 2xy^5 - 11xy^4 - 6xy^3 \). We can factor out \( xy^3 \) to make our work simpler: \[ 2xy^5 - 11xy^4 - 6xy^3 = xy^3\big(2y^2 - 11y - 6\big). \] Now we can just concentrate on factoring \( 2y^2 - 11y - 6 \). Factors of \( 2y^2: \) \( 2y \) and \( y \), \( -2y \) and \( -y \) Factors of \( -6 :\) \( -2 \) and \( 3 \), \( 2 \) and \( -3 \), \( -1 \) and \( 6 \), \( 1 \) and \(-6 \) There are many different combinations we can try for the factorization. But one thing to notice is that since the middle term is \( -11y \), we need to use factors that can get us to the number \( -11 \): \( 2y \times (-6) \) and \(y \times 1 \) would give us the middle term we want. Since we want \( 2y \) to be multiplied by \( -6 \) to give us the middle term, the \( -6 \) must exist in a different set of parentheses from \( 2y \) so that it can be distributed to and multiplied by \( 2y \). So let's check the factorization \( (2y + 1 )( y - 6 ) \). The leading term is \( 2y \cdot y =2y^2 \), and the constant term is \( 1 \cdot -6 = -6 \) as desired. The middle term is \( \big(2y \cdot (-6)\big) + 1 \cdot y = -11y \), as we wanted. So \( 2y^2 - 11y - 6 \) = \( (2y + 1 )( y - 6 ) \), implying that our final answer is \[ 2xy^5 - 11xy^4 - 6xy^3 = xy^3\big(2y^2 - 11y - 6\big) = xy^3( 2y + 1 )( y - 6 ) .\ _\square\]

Factoring polynomials in this way involves some amount of guessing and checking. You can greatly improve your speed at this process by using your number sense to figure out which combinations of numbers will successfully get you the middle term that you want.

Now that you're familiar with the different ways of factoring polynomials, let's work on some problems.

Given that \(6y^2=2014\), find the value of

\[\frac{9\big(y^4+6y^3+9y^2\big)}{y^2+6y+9}. \]

Let \(ab=1\) and define

\[ P=\frac{a}{a+1}+\frac{b}{b+1}\quad \text{ and }\quad Q=\frac{1}{a+1}+\frac{1}{b+1}.\]

What can we say about \(P\) and \(Q?\)

\[{ x }^{ 5 }-209x+56=0\]

If the above equation has two roots whose product is \(1\), find their sum.

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4.4 Solve Polynomial Equations by Factoring

Learning objectives.

• Review general strategies for factoring.
• Solve polynomial equations by factoring.
• Find roots of a polynomial function.
• Find polynomial equations given the solutions.

Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials:

• Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Determine the number of terms in the polynomial.

• Factor four-term polynomials by grouping.
• Factor trinomials (3 terms) using “trial and error” or the AC method.

Factor binomials (2 terms) using the following special products:

• Look for factors that can be factored further.
• Check by multiplying.

Note : If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization. In addition, not all polynomials with integer coefficients factor. When this is the case, we say that the polynomial is prime.

If an expression has a GCF, then factor this out first. Doing so is often overlooked and typically results in factors that are easier to work with. Furthermore, look for the resulting factors to factor further; many factoring problems require more than one step. A polynomial is completely factored when none of the factors can be factored further.

Factor: 54 x 4 − 36 x 3 − 24 x 2 + 16 x .

This four-term polynomial has a GCF of 2 x . Factor this out first.

54 x 4 − 36 x 3 − 24 x 2 + 16 x = 2 x ( 27 x 3 − 18 x 2 − 12 x + 8 )

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

Answer: 2 x ( 3 x − 2 ) 2 ( 3 x + 2 ) . The check is left to the reader.

Factor: x 4 − 3 x 2 − 4 .

This trinomial does not have a GCF.

x 4 − 3 x 2 − 4 = ( x 2             ) ( x 2             ) = ( x 2 + 1 ) ( x 2 − 4 )          D i f f e r e n c e   o f   s q u a r e s = ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

The factor ( x 2 + 1 ) is prime and the trinomial is completely factored.

Answer: ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

Factor: x 6 + 6 x 3 − 16 .

Begin by factoring x 6 = x 3 ⋅ x 3 and look for the factors of 16 that add to 6.

The factor ( x 3 − 2 ) cannot be factored any further using integers and the factorization is complete.

Answer: ( x 3 − 2 ) ( x + 2 ) ( x 2 + 2 x + 4 )

Try this! Factor: 9 x 4 + 17 x 2 − 2

Answer: ( 3 x + 1 ) ( 3 x − 1 ) ( x 2 + 2 )

Solving Polynomial Equations by Factoring

In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property A product is equal to zero if and only if at least one of the factors is zero. :

a ⋅ b = 0    if and only if    a = 0  or  b = 0

The zero-product property is true for any number of factors that make up an equation. In other words, if any product is equal to zero, then at least one of the variable factors must be equal to zero. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Solve: 2 x ( x − 4 ) ( 5 x + 3 ) = 0 .

Set each variable factor equal to zero and solve.

2 x = 0   or   x − 4 = 0   or   5 x + 3 = 0 2 x 2 = 0 2 x = 4 5 x 5 = − 3 5 x = 0 x = − 3 5

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

Answer: The solutions are 0, 4, and − 3 5 .

Of course, most equations will not be given in factored form.

Solve: 4 x 3 − x 2 − 100 x + 25 = 0 .

Begin by factoring the left side completely.

4 x 3 − x 2 − 100 x + 25 = 0 F a c t o r   b y   g r o u p i n g . x 2 ( 4 x − 1 ) − 25 ( 4 x − 1 ) = 0 ( 4 x − 1 ) ( x 2 − 25 ) = 0 F a c t o r   a s   a   d i f f e r e n c e   o f   s q u a r e s . ( 4 x − 1 ) ( x + 5 ) ( x − 5 ) = 0

Set each factor equal to zero and solve.

4 x − 1 = 0 or x + 5 = 0 or x − 5 = 0 4 x = 1 x = − 5 x = 5 x = 1 4

Answer: The solutions are 1 4 , −5, and 5.

Using the zero-product property after factoring an equation that is equal to zero is the key to this technique. However, the equation may not be given equal to zero, and so there may be some preliminary steps before factoring. The steps required to solve by factoring The process of solving an equation that is equal to zero by factoring it and then setting each variable factor equal to zero. are outlined in the following example.

Solve: 15 x 2 + 3 x − 8 = 5 x − 7 .

Step 1: Express the equation in standard form, equal to zero. In this example, subtract 5 x from and add 7 to both sides.

15 x 2 + 3 x − 8 = 5 x − 7 15 x 2 − 2 x − 1 = 0

Step 2: Factor the expression.

( 3 x − 1 ) ( 5 x + 1 ) = 0

Step 3: Apply the zero-product property and set each variable factor equal to zero.

3 x − 1 = 0         or         5 x + 1 = 0

Step 4: Solve the resulting linear equations.

3 x − 1 = 0 or 5 x + 1 = 0 3 x = 1 5 x = − 1 x = 1 3 x = − 1 5

Answer: The solutions are 1 3 and − 1 5 . The check is optional.

Solve: ( 3 x + 2 ) ( x + 1 ) = 4 .

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to 4. However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

( 3 x + 2 ) ( x + 1 ) = 4 3 x 2 + 3 x + 2 x + 2 = 4 3 x 2 + 5 x + 2 = 4 3 x 2 + 5 x − 2 = 0

Once it is in standard form, we can factor and then set each factor equal to zero.

                ( 3 x − 1 ) ( x + 2 ) = 0 3 x − 1 = 0               or           x + 2 = 0               3 x = 1                                                 x = − 2                     x = 1 3

Answer: The solutions are 1 3 and −2.

Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0

A root A value in the domain of a function that results in zero. of a function is a value in the domain that results in zero. In other words, the roots occur when the function is equal to zero, f ( x ) = 0 .

Find the roots: f ( x ) = ( x + 2 ) 2 − 4 .

To find roots we set the function equal to zero and solve.

f ( x ) = 0 ( x + 2 ) 2 − 4 = 0 x 2 + 4 x + 4 − 4 = 0 x 2 + 4 x = 0 x ( x + 4 ) = 0

Next, set each factor equal to zero and solve.

x = 0   or     x + 4 = 0 x = − 4

We can show that these x -values are roots by evaluating.

f ( 0 ) = ( 0 + 2 ) 2 − 4         f ( − 4 ) = ( − 4 + 2 ) 2 − 4 = 4 − 4   = ( − 2 ) 2 − 4 = 0         ✓ = 4 − 4 = 0         ✓

Answer: The roots are 0 and −4.

If we graph the function in the previous example we will see that the roots correspond to the x -intercepts of the function. Here the function f is a basic parabola shifted 2 units to the left and 4 units down.

Find the roots: f ( x ) = x 4 − 5 x 2 + 4 .

f ( x ) = 0 x 4 − 5 x 2 + 4 = 0 ( x 2 − 1 ) ( x 2 − 4 ) = 0 ( x + 1 ) ( x − 1 ) ( x + 2 ) ( x − 2 ) = 0

x + 1 = 0         or         x − 1 = 0         or         x + 2 = 0         or         x − 2 = 0 x = − 1 x = 1 x = − 2 x = 2

Answer: The roots are −1, 1, −2, and 2.

Graphing the previous function is not within the scope of this course. However, the graph is provided below:

Notice that the degree of the polynomial is 4 and we obtained four roots. In general, for any polynomial function with one variable of degree n , the fundamental theorem of algebra Guarantees that there will be as many (or fewer) roots to a polynomial function with one variable as its degree. guarantees n real roots or fewer. We have seen that many polynomials do not factor. This does not imply that functions involving these unfactorable polynomials do not have real roots. In fact, many polynomial functions that do not factor do have real solutions. We will learn how to find these types of roots as we continue in our study of algebra.

Find the roots: f ( x ) = − x 2 + 10 x − 25 .

f ( x ) = 0 − x 2 + 10 x − 25 = 0 − ( x 2 − 10 x + 25 ) = 0 − ( x − 5 ) ( x − 5 ) = 0

Next, set each variable factor equal to zero and solve.

x − 5 = 0 or x − 5 = 0 = 5 x = 5

A solution that is repeated twice is called a double root A root that is repeated twice. . In this case, there is only one solution.

Answer: The root is 5.

The previous example shows that a function of degree 2 can have one root. From the factoring step, we see that the function can be written

f ( x ) = − ( x − 5 ) 2

In this form, we can see a reflection about the x -axis and a shift to the right 5 units. The vertex is the x -intercept, illustrating the fact that there is only one root.

Try this! Find the roots of f ( x ) = x 3 + 3 x 2 − x − 3 .

Answer: ±1, −3

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x , where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 40 feet.

We are asked to find the speed x where the safe stopping distance d ( x ) = 40 feet.

d ( x ) = 40 1 20 x 2 + x = 40

To solve for x , rewrite the resulting equation in standard form. In this case, we will first multiply both sides by 20 to clear the fraction.

20 ( 1 20 x 2 + x ) = 20 ( 40 ) x 2 + 20 x = 800 x 2 + 20 x − 800 = 0

Next factor and then set each factor equal to zero.

x 2 + 20 x − 800 = 0 ( x + 40 ) ( x − 20 ) = 0 x + 40 = 0 o r x − 20 = 0 x = − 40 x = 20

The negative answer does not make sense in the context of this problem. Consider x = 20 miles per hour to be the only solution.

Answer: 20 miles per hour

Finding Equations with Given Solutions

We can use the zero-product property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Find a quadratic equation with integer coefficients, given solutions − 3 2 and 1 3 .

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

x = − 3 2 or x = 1 3 2 x = − 3 3 x = 1 2 x + 3 = 0 3 x − 1 = 0

The product of these linear factors is equal to zero when x = − 3 2 or x = 1 3 .

( 2 x + 3 ) ( 3 x − 1 ) = 0

Multiply the binomials and present the equation in standard form.

6 x 2 − 2 x + 9 x − 3 = 0 6 x 2 + 7 x − 3 = 0

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

Answer: 6 x 2 + 7 x − 3 = 0

Find a polynomial function with real roots 1, −2, and 2.

Given solutions to f ( x ) = 0 we can find linear factors.

x = 1 or x = − 2 or x = 2 x − 1 = 0 x + 2 = 0 x − 2 = 0

Apply the zero-product property and multiply.

( x − 1 ) ( x + 2 ) ( x − 2 ) = 0 ( x − 1 ) ( x 2 − 4 ) = 0 x 3 − 4 x − x 2 + 4 = 0 x 3 − x 2 − 4 x + 4 = 0

Answer: f ( x ) = x 3 − x 2 − 4 x + 4

Try this! Find a polynomial equation with integer coefficients, given solutions 1 2 and − 3 4 .

Answer: 8 x 2 + 2 x − 3 = 0

Key Takeaways

• Factoring and the zero-product property allow us to solve equations.
• To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting equations are the solutions to the original.
• Not all polynomial equations can be solved by factoring. We will learn how to solve polynomial equations that do not factor later in the course.
• A polynomial function can have at most a number of real roots equal to its degree. To find roots of a function, set it equal to zero and solve.
• To find a polynomial equation with given solutions, perform the process of solving by factoring in reverse.

Topic Exercises

Part a: general factoring.

Factor completely.

50 x 2 − 18

12 x 3 − 3 x

10 x 3 + 65 x 2 − 35 x

15 x 4 + 7 x 3 − 4 x 2

6 a 4 b − 15 a 3 b 2 − 9 a 2 b 3

8 a 3 b − 44 a 2 b 2 + 20 a b 3

36 x 4 − 72 x 3 − 4 x 2 + 8 x

20 x 4 + 60 x 3 − 5 x 2 − 15 x

3 x 5 + 2 x 4 − 12 x 3 − 8 x 2

10 x 5 − 4 x 4 − 90 x 3 + 36 x 2

x 4 − 23 x 2 − 50

2 x 4 − 31 x 2 − 16

− 2 x 5 − 6 x 3 + 8 x

− 36 x 5 + 69 x 3 + 27 x

54 x 5 − 78 x 3 + 24 x

4 x 6 − 65 x 4 + 16 x 2

x 6 − 7 x 3 − 8

x 6 − 25 x 3 − 54

3 x 6 + 4 x 3 + 1

27 x 6 − 28 x 3 + 1

Part B: Solving Polynomial Equations by Factoring

( 6 x − 5 ) ( x + 7 ) = 0

( x + 9 ) ( 3 x − 8 ) = 0

5 x ( 2 x − 5 ) ( 3 x + 1 ) = 0

4 x ( 5 x − 1 ) ( 2 x + 3 ) = 0

( x − 1 ) ( 2 x + 1 ) ( 3 x − 5 ) = 0

( x + 6 ) ( 5 x − 2 ) ( 2 x + 9 ) = 0

( x + 4 ) ( x − 2 ) = 16

( x + 1 ) ( x − 7 ) = 9

( 6 x + 1 ) ( x + 1 ) = 6

( 2 x − 1 ) ( x − 4 ) = 39

x 2 − 15 x + 50 = 0

x 2 + 10 x − 24 = 0

3 x 2 + 2 x − 5 = 0

2 x 2 + 9 x + 7 = 0

1 10 x 2 − 7 15 x − 1 6 = 0

1 4 − 4 9 x 2 = 0

6 x 2 − 5 x − 2 = 30 x + 4

6 x 2 − 9 x + 15 = 20 x − 13

5 x 2 − 23 x + 12 = 4 ( 5 x − 3 )

4 x 2 + 5 x − 5 = 15 ( 3 − 2 x )

( x + 6 ) ( x − 10 ) = 4 ( x − 18 )

( x + 4 ) ( x − 6 ) = 2 ( x + 4 )

4 x 3 − 14 x 2 − 30 x = 0

9 x 3 + 48 x 2 − 36 x = 0

1 3 x 3 − 3 4 x = 0

1 2 x 3 − 1 50 x = 0

− 10 x 3 − 28 x 2 + 48 x = 0

− 2 x 3 + 15 x 2 + 50 x = 0

2 x 3 − x 2 − 72 x + 36 = 0

4 x 3 − 32 x 2 − 9 x + 72 = 0

45 x 3 − 9 x 2 − 5 x + 1 = 0

x 3 − 3 x 2 − x + 3 = 0

x 4 − 5 x 2 + 4 = 0

4 x 4 − 37 x 2 + 9 = 0

Find the roots of the given functions.

f ( x ) = x 2 + 10 x − 24

f ( x ) = x 2 − 14 x + 48

f ( x ) = − 2 x 2 + 7 x + 4

f ( x ) = − 3 x 2 + 14 x + 5

f ( x ) = 16 x 2 − 40 x + 25

f ( x ) = 9 x 2 − 12 x + 4

g ( x ) = 8 x 2 + 3 x

g ( x ) = 5 x 2 − 30 x

p ( x ) = 64 x 2 − 1

q ( x ) = 4 x 2 − 121

f ( x ) = 1 5 x 3 − 1 x 2 − 1 20 x + 1 4

f ( x ) = 1 3 x 3 + 1 2 x 2 − 4 3 x − 2

g ( x ) = x 4 − 13 x 2 + 36

g ( x ) = 4 x 4 − 13 x 2 + 9

f ( x ) = ( x + 5 ) 2 − 1

g ( x ) = − ( x + 5 ) 2 + 9

f ( x ) = − ( 3 x − 5 ) 2

g ( x ) = − ( x + 2 ) 2 + 4

Given the graph of a function, determine the real roots.

The sides of a square measure x − 2 units. If the area is 36 square units, then find x .

The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

The profit in dollars generated by producing and selling n bicycles per week is given by the formula P ( n ) = − 5 n 2 + 400 n − 6000 . How many bicycles must be produced and sold to break even?

The height in feet of an object dropped from the top of a 64-foot building is given by h ( t ) = − 16 t 2 + 64 where t represents the time in seconds after it is dropped. How long will it take to hit the ground?

A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

What is the length of each side of the cardboard sheet if the volume of the box is to be 98 cubic inches?

The height of a triangle is 4 centimeters less than twice the length of its base. If the total area of the triangle is 48 square centimeters, then find the lengths of the base and height.

A uniform border is to be placed around an 8 × 10 inch picture.

If the total area including the border must be 168 square inches, then how wide should the border be?

The area of a picture frame including a 3-inch wide border is 120 square inches.

If the width of the inner area is 2 inches less than its length, then find the dimensions of the inner area.

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 75 feet.

A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula R ( n ) = 12 n − 0.6 n 2 where n represents the number of palettes of product sold ( 0 ≤ n < 20 ) . Determine the number of palettes sold in a day if the revenue was 45 thousand dollars.

Part C: Finding Equations with Given Solutions

Find a polynomial equation with the given solutions.

Find a function with the given roots.

2 5 , − 1 3

5 double root

−3 double root

Recall that if | X | = p , then X = − p or X = p . Use this to solve the following absolute value equations.

| x 2 − 8 | = 8

| 2 x 2 − 9 | = 9

| x 2 − 2 x − 1 | = 2

| x 2 − 8 x + 14 | = 2

| 2 x 2 − 4 x − 7 | = 9

| x 2 − 3 x − 9 | = 9

Part D: Discussion Board

Explain to a beginning algebra student the difference between an equation and an expression.

What is the difference between a root and an x -intercept? Explain.

Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.

Research and discuss the fundamental theorem of algebra.

2 ( 5 x + 3 ) ( 5 x − 3 )

5 x ( x + 7 ) ( 2 x − 1 )

3 a 2 b ( 2 a + b ) ( a − 3 b )

4 x ( x − 2 ) ( 3 x + 1 ) ( 3 x − 1 )

x 2 ( 3 x + 2 ) ( x + 2 ) ( x − 2 )

( x 2 + 2 ) ( x + 5 ) ( x − 5 )

• − 2 x ( x 2 + 4 ) ( x − 1 ) ( x + 1 )

6 x ( x + 1 ) ( x − 1 ) ( 3 x + 2 ) ( 3 x − 2 )

( x + 1 ) ( x 2 − x + 1 ) ( x − 2 ) ( x 2 + 2 x + 4 )

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

0, 5 2 , − 1 3

− 1 2 , 1, 5 3

− 5 3 , 1 2

0, − 3 2 , 5

± 1 3 , 1 5

−3, −1, 0, 2

20 or 60 bicycles

30 miles per hour

x 2 − 2 x − 15 = 0

3 x 2 − 7 x + 2 = 0

x 2 + 4 x = 0

x 2 − 49 = 0

x 3 − x 2 − 9 x + 9 = 0

f ( x ) = 6 x 2 − 7 x + 2

f ( x ) = 16 x 2 − 9

f ( x ) = x 2 − 10 x + 25

f ( x ) = x 3 − 2 x 2 − 3 x

Answer may vary

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Solving & Factoring Polynomials: Examples

Solving Factoring Examples

These exercises can be very long, so I've only shown three examples so far. Here are a few more, for practice:

Find the real-number solutions to x 6 + 9 x 5 + 11 x 4 − 22 x 3 − 9 x 2 − 11 x + 21 = 0 .

They've given me an equation, and have asked for the solutions to that equation. So I'll be finding x -values, rather than factors.

First, I'll try the trick with 1 and −1 . Trying x  = 1 , I get:

1 + 9 + 11 − 22 − 9 − 11 + 21 = 0

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Excellent! So x  = 1 is one of the zeroes. Trying x  = −1 , I get:

1 − 9 + 11 + 22 − 9 + 11 + 21 = 48

Okay; so that one isn't a zero. But, to reduce my polynomial by the one factor corresponding to this zero, I'll do my first synthetic division:

So my reduced polynomial is equation is:

x 5 + 10 x 4 + 21 x 3 − x 2 − 10 x − 21 = 0

This is so nasty... I'm gonna try the trick with x  = 1 again, just in case it's a root twice:

1 + 10 + 21 − 1 − 10 − 21 = 0

Nice! Okay; here's my second synthetic division:

Alright. My new polynomial equation now is:

x 4 + 11 x 3 + 32 x 2 + 31 x + 21 = 0

All the coefficients are positive, so +1 cannot be a zero again. Now it's time for the Rational Roots Test:

x = ±1, 3, 7, 21

Hm... I've already scratched off ±1 directly. Because all the coefficients are positive, then I know that +3, +7, and +21 are out, too. I'd rather stay small, if I can, so I'll try −3 next:

So now my polynonial equation is:

x 3 + 8 x 2 + 8 x + 7 = 0

From this reduced polynomial, I can see that I can cross −21 and −3 off the list of possible roots; they clearly won't work in this reduced polynomial. So I guess I'm left with −7 :

And now I'm down to a quadratic, which I can easily solve:

x 2 + x + 1 = 0

x = (−1 ± sqrt[−3])/2

Because there's a minus inside the radical, I know that the solutions of this quadratic are complex numbers; it has no real zeroes. Since they asked for just the real-valued roots of the original polynomial, then I can ignore these last two zeroes. Then my answer is:

x = 1, −3, −7

I didn't check the graph when I did the above, but it does confirm my answer:

The intercepts at x  = −7 and at x  = −3 are clear. The intercept at x  = 1 is clearly repeated, because of how the curve bounces off the x -axis at this point, and goes back the way it came.

Note: This polynomial's graph is so steep in places that it sometimes disappeared in my graphing software. I had to fiddle with the axis values and window size to get the whole curve to show up. When using your calculator, don't stick only with the default screen for graphs; play with the axis values until you get a picture that's useful.

Factor completely: 2 x 5 − 3 x 4 − 9 x 3 + 3 x 2 − 11 x + 6

They've given me an expression rather than an equation, and have told me to factor. So I'll be finding factors rather than x -values, and I'll need to keep track of everything I pull out, from beginning to end.

There is no factor common to all terms, so there is nothing to pull out yet. I'll check for zeroes of the associated polynomial equation (setting the original expression equal to zero), and see what I can find. Then I'll convert the zeroes to factors, and pull them out.

First, I'll try the usual shortcut with ±1 ; the positive first:

2 − 3 − 9 + 3 − 11 + 6 = −12

No joy. I'll try the negative now:

−2 − 3 + 9 + 3 + 11 + 6 = 90

That's even worse. Okay, now I'll use the Rational Roots Test to create a list of maybe-solution values:

±(1, 2, 3, 6)/(1, 2)

= ±1/2, 1, 3/2, 2, 3, 6

I already know that I can ignore ±1 . Descartes' Rule of Signs tells me that there will be four, two, or zero positive roots; and one (definite) negative root. So I'll start with the negative integers:

Okay; I've found that x  = −2 is a root, which means that x  + 2 is a factor. Also, I've reduced the expression still needing to be factored to:

2 x 4 − 7 x 3 + 5 x 2 − 7 x + 3

The constant term is 3 , so I know that ±2 cannot be a solution to what's left, nor can ±6 . Also, I've already found the one negative zero. So this leaves me with:

1/2, 3/2, 3

I'm trying to avoid fractions, so I'll try the last integer possibility:

The last row above is a four-term polynomial that looks like it can be factored in pairs:

(2 x 3 − x 2 ) + (2 x − 1)

x 2 (2 x − 1) + 1(2 x − 1)

(2 x − 1)( x 2 + 1)

The quadratic factor is the sum of squares, so it isn't factorable. This means I'm done, and my complete factorization is:

( x  + 2)​( x  − 3)​(2 x  − 1)​( x 2  + 1)

The method for answering the two exercises above is the method that I learned, back in the olden times when dinosaurs ruled the world and calculators were made with bear skins and stone knives. It's probably at least similar to the method that you've seen in your book, and your instructor likely expects you do show work along the lines of what I did above.

However, if you have a graphing calculator (and nearly everybody does, nowadays), you can avoid wasting quite so much time on maybe-solution values that turn out not to work.

Find all zeroes of y = 8 x 5 − 58 x 4 + 137 x 3 − 118 x 2 + 33 x + 18

Before doing anything else, I'll do a quick graph:

Looking at the graph, I know to check x  = 3 twice:

The original polynomial was degree-five. I've found one zero of multiplicity two, which leaves at most three more zeroes. Looking at the polynomial represented by the last row above, the Rational Roots Test says that any remaining "nice" zeroes will be among these:

±(1, 2)/(1, 2, 4, 8)

= ±1/8, 1/4, 1/2, 1, 2

Now I can see that there's a common factor of four that can be divided out and discarded, leaving me with:

2 x 2 − 3 x + 2 = 0

x = (−(−3) ± sqrt[(−3) 2 − 4(2)(2)])/(2(2))

= (3 ± sqrt[9 − 16])/4

= (3 ± sqrt[−7])/4

They asked me for all of the zeroes, not just all of the real-valued ones, so I have to include these two roots that don't show up on the graph. However, by checking the graph first, I was able to save a lot of time in arriving at my answer:

x = −(1/4), 3, (3 ± sqrt[−7])/4

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There is a variant of these exercises, where they provide one or more factors (of an expression) or zeroes (of an equation or function), and they want you to find the rest of them. I've got examples of how this works in the last page of the lesson on synthetic division. Varient exercises are often a bit messier and, to answer them, you're expected to have a deeper understanding of how the Quadratic Formula generates solutions in pairs, because of the " ± ". Otherwise, they work in pretty much the same way.

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Factoring Polynomials

When numbers are multiplied together, each of the numbers multiplied to get the product is called a factor . Sometimes it is desirable to write a polynomial as the product of certain of its factors. This operation is called factoring . Here we are interested in factoring polynomials with integral coefficients.

A polynomial is said to be factored completely if it is expressed as the product of polynomials with integral coefficients, and no one of the factors can still be written as the product of two polynomials with integral coefficients.

Following is a discussion of factoring some special polynomials.

Factors Common to All Terms

The greatest common factor (GCF) of a set of integers is defined as the greatest integer that divides each number of that set of integers.

The GCF can be obtained as follows:

1.  Factor the integers into their prime factors.

2.   Write the factors in the exponent form.

3.   Take the common bases each to its lowest exponent.

Find the GCF of 30 , 45 , 60 .

60 = 2^2*3*5

The common bases are 3 and 5 .

The least exponent of 3 is 1 and of 5 is 1 .

Hence the GCF = 3^1*5^1 = 15 .

The greatest common factor of a set of monomials can be found by taking the product of the GCF of the coefficients of the monomials and the common literal bases. each to its lowest exponent.

Find the GCF of 9x^(3)y^(2) , 12x^4y , -15x^5 .

9x^3y^2=3^2x^3y^2

12x^4y=2^2*3x^4y

  -15x^5=-3*5x^5

The common bases are 3 and x .

The least exponent of 3 is 1 and of x is 3 .

Hence the GCF = 3x^3 .

Find the GCF of 6a^4(x - y)^2 , 9a^3(x - y)^3 , 12a^2(x - y)^4 .

6a^4(x - y)^2=2*3a^4(x-y)^2

9a^3(x - y)^3=3^(2)a^(3)(x-y)^3

12a^2(x - y)^4=2^2*3a^2(x-y)^4

The common bases are 3 , a , and (x - y) .

The least exponent of 3 is 1 , of a is 2 , and of (x - y) is 2 .

Hence the GCF = 3a^2(x - y)^2 .

Note Since (1 - x)=-(x - 1) , the GCF of a(x - 1) , b(1 - x) is either (x - 1) or (1-x) .

When the terms of a polynomial have a common factor, the distributive law,

ab_1+ab_2+ab_3+...+ab_n=a(b_1+b_2+b_3+...b_n)

is used to factor the polynomial. One factor is the greatest common factor of all the terms of the polynomial. The other factor is the entire quotient, obtained by dividing each term of the polynomial by the common factor; that is,

ab_1+ab_2+ab_3+...+ab_n=a((ab_1)/a+(ab_2)/a+(ab_3)/a...+(ab_n)/a)

= a(b_1+b_2+b_3+...b_n)

Factor the expression 3a^2 - a .

The greatest common factor is a .

3a^2 - a = a((3a^2)/a-a/a)

            = a(3a-1)

Factor the polynomial 6x^(3)y^(2) + 12x^(2)y^(2) - 24xy^(2) .

The greatest common factor is 6xy^2 .

6x^(3)y^(2) + 12x^(2)y^(2) - 24xy^(2) = 6xy^2((6x^(3)y^(2))/(6xy^2)+(12x^(2)y^(2))/(6xy^2)-(24xy^2)/(6xy^2)

                                         = 6xy^2(x^2+2x-4)

Factor the polynomial 4x^2(2x - 1) - 8x(2x - 1)^2 .

The greatest common factor is 4x(2x - 1) .

4x^2(2x - 1) - 8x(2x - 1)^2 = 4x(2x - 1)[(4x^2(2x-1))/(4x(2x-1))-(8x(2x-1)^2)/(4x(2x-1))

                                             = 4x(2x-1)[x-2(2x-1)

                                             = 4x(2x-1)[x-4x+2

                                             = 4x(2x-1)(2-3x)

This is how our factorization calculator solves the problem above. You can see similar problems solved by clicking on 'Solve similar' button.

Factoring a Binomial

The methods of factoring polynomials will be presented according to the number of terms in the polynomial to be factored.

A monomial is already in factored form; thus the first type of polynomial to be considered for factoring is a binomial. Here we shall discuss factoring one type of binomials.

Squares and Square Roots

The squares of the numbers      3 ,        5^2 ,        a ,        x^2 ,        and        b^3

are, respectively,                      3^2 ,      5^4 ,       a^2 ,       x^4 ,         and      b^6

The 3 , 5^2 , a , x^2 , and b^3 are called the square roots of 3^2 , 5^4 , a^2 , x^4 , and b^6 , respectively.

Although the square of both (+ 3) and (- 3) is 9 , when we talk about the square root of 9 , we will mean the positive number 3 and not the negative number (-3) .

A number is said to be a perfect square if its square root is a rational number.

The square root of a specific number can be found by factoring the number into its prime factors, writing it in the exponent form, and then taking each base to one-half of its original exponent (when we square a number, we multiply its exponent by 2 ).

1.  root(64) = root(2^6) = 2^3 = 8

2.   root(144) = root(2^(4)*3^(2) = 2^2*3^1 = 12

When a is a literal number and n {is-in} N , we define root(a^(2n)) as (root(a))^(2n) = a^n . If the exponent is not divisible by 2 , the number is not a perfect square.

1.  root(a^4) = a^2

2.   root(x^(2)y^(6)) = xy^3

3.   root(4x^(2)y^(4)) = 2xy^2

The numbers 2 , 3 , 5 , 7 , 8 , 10 etc., are not perfect square numbers. This means there are no rational numbers whose squares are 2 , 3 , 5 etc.

The square roots of numbers that are not perfect squares are called irrational numbers .

Difference of Two Squares

The product of the two factors (a + b)(a - b) is a^2 - b^2 , the difference of two perfect square terms. The factors of the difference of two squares are the sum and difference of the respective square roots of the two squares.

Factor 9a^2-4

The square root of 9a^2 is 3a and of 4 is 2 . Hence 9a^2 - 4=(3a + 2)(3a - 2)

Note Remember to factor the polynomial completely.

Factor completely x^4 - 81y^4 .

x^4 - 81y^4 = (x^2+9y^2)(x^2-9y^2)

               = (x^2+9y^2)(x+3y)(x-3y)

Note Before checking if the binomial is a difference of two squares, check for a common factor. That is always the first operation to be performed.

Factor completely 6x^4 - 6 .

6x^4 - 6 = 6(x^4-1)

            = 6(x^2+1)(x^2-1)

            = 6(x^2+1)(x+1)(x-1)

Note (a + b)(a - b) = (a - b)(a + b)

Factoring a Trinomial

Factoring trinomials is divided into two cases:

1.   When the trinomial is of the form x^2 + bx + c , b , c ∈ I , b!=0 , c!=0 .

2.   When the trinomial is of the form ax^2 + bx + c , a!=1 , a , b , c ∈ I , b!=0 , c!=0 .

Trinomials of the Form , b, c ∈ I , and b ≠ 0, c ≠ 0

Consider the following products:

      (x+m)(x+n) = x^2+(m+n)x+mn

      (x-m)(x-n) = x^2 + (-m - n)x + mn

      (x+m)(x-n) = x^2+(m-n)x-mn

      (x-m)(x+n) = x^2 + (-m + n)x - mn

We note the following relations between the products and their factors:

1.  The first term in each factor is the square root of the square term in the trinomial.

2.   The product of the second terms of the factors is the third term in the trinomial.

3.  The sum of the second terms, signed numbers, is the coefficient of the middle term in the trinomial.

Note    To find the second terms in the factors, look for two signed numbers whose product is the third term in the trinomial and whose sum is the coefficient of the middle term in the trinomial.

Note    When the sign of the third term in the trinomial is plus, the two signed numbers have like signs and are the same as the sign of the middle term in the trinomial.

Note   When the sign of the third term in the trinomial is minus, the two signed numbers have different signs, and the larger one numerically has the sign of the middle term in the trinomial.

Factor x^2+8x+15

The first term of each factor is root(x^2) = x

Hence  x^2+8x+15 = ( x           )( x            )

Since the sign of the last term (+15) is plus, the two signed numbers in the factors have like signs.

Since the sign of the middle term (+8x) is plus, the two signed numbers are positive.

x^2+8x+15 = ( x +          )( x +           )

We look for two natural numbers whose product is 15 and whose sum is 8 . The two numbers are 3 and 5 .

Hence x^2+8x+15 = (x+3)(x+5)

Factor x^2-10x+24

Hence x^2-10x+24 = ( x           )( x            )

Since the sign of the last term (+24) is plus, the two signed numbers in the factors have like signs.

Since the sign of the middle term (-10x) is minus, the two signed numbers are negative.

x^2-10x+24 = ( x -          )( x -           )

We look for two natural numbers whose product is 24 and whose sum is 10 . The two numbers are 4 and 6 .

Hence x^2-10x+24 = (x-4)(x-6)

Factor x^2-5x-36

x^2-5x-36 = ( x           )( x            )

Since the sign of the last term (-36) is minus, the two numbers in the factors have different signs.

x^2-5x-36 = ( x +          )( x -          )

Since the sign of the middle term (-5x) is minus, the numerically larger number has the negative sign.

x^2-5x-36 = ( x + smaller number)( x - larger number)

We look for two natural numbers whose product is 36 and whose difference is 5 . The two numbers are 4 and 9 .

Hence x^2-5x-36 = (x + 4)(x - 9)

You can check below how our factorization calculator factorize the trinomial above. You can see similar problems solved by clicking on 'Solve similar' button.

Factor x^2+3x-28

x^2+3x-28 = ( x           )( x            )

Since the sign of the last term (-28) is minus, the two numbers in the factors have different signs.

x^2+3x-28 = ( x +          )( x -           )

Since the sign of the middle term (+3x) is plus, the numerically larger number has the plus sign.

x^2+3x-28 = ( x + larger number)( x - smaller number)

We look for two natural numbers whose product is 28 and whose difference is 3 . The two numbers are 4 and 7 .

Hence x^2+3x-28 = (x+7)(x-4)

Factor (x - y)^2 - 3(x - y) - 10 .

(x - y)^2 - 3(x - y) - 10 is of the form a^2 - 3a - 10 , whose factors are (a - 5)(a + 2) .

Hence (x - y)^2 - 3(x - y) - 10 = [(x - y) - 5][(x - y) + 2

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

Note       When the third term of the trinomial is a large number and its factors are not obvious, write the number as the product of its prime factors; then make products

              of factors using combinations of the primes.

Note       (x + a)(x + b) = (x + b)(x + a)

Trinomials of the Form , a ≠ 1, a, b, c ∈ I , b ≠ 0, c ≠ 0

Consider the product

     (2x + 4)(x + 3) = 2x^2 + 10x + 12

The first factor on the left contains the common factor 2 :

     2x + 4 = 2(x + 2)

Also, the expanded product contains the common factor 2 :

     2x^2 + 10x +12 = 2(x^2 + 5x + 6)

In general, if a factor of a product contains a common factor, then the expanded product will also contain that common factor.

On the other hand, if no factor in a product, (x + 5)(3x - 2) , contains a common factor, then the expanded product, 3x^2 + 13x - 10 , will not have a common factor. Conversely, if the terms of a product do not have a common factor, then neither will any of its factors.

In order to learn how to factor a trinomial of the form ax^2 + bx + c , let us first look at how we multiply two factors together to get a product of this form.

We multiply (2x + 3)(4x - 5) as follows:

Let us go over the same multiplication again, as shown in Figure 6.1.

The sum of the products in the direction of the arrows,

is the middle term of the trinomial.

The following example illustrates how to use the scissors in factoring a trinomial

ax^2+bx+c , a!=1 , a , b , c ∈ I .

Factor 6x^2-5x-6 .

Find all possible pairs of factors whose product is the first term of , the trinomial; each factor must contain the square root of the literal number. Write these factors at the left side of the scissors.

Find all possible pairs of factors whose product is the third term of the trinomial, disregarding the signs, and write them at the right side of the scissors.

Write all possible arrangements using the factors of the first term and the factors of the third term.

The eight scissors just shown give all possible arrangements of the factors of the first term of the trinomial and the factors of the third term of the trinomial.

The terms on top of the scissors form the first factor of the product, and the terms on the bottom of the scissors form the second factor of the product.

Since there is no common factor in the trinomial. there should not be a common factor between the terms at the top of the scissors or a common factor between the terms at the bottom of the scissors. If there is a common factor between the terms at the top or the terms at the bottom in an arrangement. that arrangement cannot be the correct one. Arrangements (1) , (3) , (4) , (5) , (6) and (7) have common factors, and we eliminate them.

The candidates are now limited to two arrangements.

The middle term of the trinomial. which is the sum of the products in the direction of the arrows. will indicate which arrangement is the correct one.

Since the first arrangement gives x and 36x for the middle term, which cannot give a sum of -5x , the first arrangement is not the correct one. The second arrangement gives 9x and 4x for the middle term, and by taking the 9x with a minus sign and 4x with a plus sign, we get -9x + 4x = -5x .

Hence the correct arrangement is

The factors of the first term of the trinomial are always taken positive. Thus, in order to arrive at -9x , the 3 at the right side of the scissors has to be taken negative while the 2 has to be taken positive to arrive at +4x . The complete arrangement is

Hence 6x^2 - 5x - 6 = (2x - 3)(3x + 2)

Note       When the trinomial has a common factor, factor it first before you attempt factoring by the scissors.

Note       There is no reason to write any arrangement with a common factor between the top terms or a common factor between the bottom terms.

Note       When the coefficient of the first term, or the third term of the trinomial is a large number, write the number as the product of its prime factors, and form products of factors using combinations of the primes.

See how our factorization calculator solve the trinomial above. You can see similar problems solved by clicking on 'Solve similar' button.

Factor 6x^2+19x+15 .

Hence  6x^2+19x+15 = (2x+3)(3x+5)

Factor 36x^4-241x^2+100 .

Hence  36x^4-241x^2+100 = (4x^2-25)(9x^2-4)

                                          = (2x+5)(2x-5)(3x+2)(3x-2)

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