problem solving involving polynomial equation example

Study Guides > College Algebra

Solve real-world applications of polynomial equations.

We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section.

Example 8: Solving Polynomial Equations

A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\frac{1}{3}w[/latex]. Let’s write the volume of the cake in terms of width of the cake.

Substitute the given volume into this equation.

Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351[/latex], and [latex]\pm 1053[/latex]. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[/latex].

Synthetic Division with divisor = 1, and quotient = {1, 4, 0, -1053}. Solution is {1, 5, 5, -1048}

Since 1 is not a solution, we will check [latex]x=3[/latex].

Since 3 is not a solution either, we will test [latex]x=9[/latex].

Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.

The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?

Licenses & Attributions

Cc licensed content, shared previously.

Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution . License terms: Download For Free at : http://cnx.org/contents/ [email protected] ..

Please add a message.

Message received. Thanks for the feedback.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Course: Algebra 2 > Unit 5

Graphs of polynomials

Graphs of polynomials: Challenge problems

Polynomial graphs: FAQ

What you should be familiar with before taking this lesson

End behavior of polynomials
Zeros of polynomials and their graphs

What you will do in this lesson

(Choice A) A
(Choice B) B
(Choice C) C
(Choice D) D

As x → + ∞ ‍ , y → − ∞ ‍ and as x → − ∞ ‍ , y → + ∞ ‍ .

Want to join the conversation?

Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page

Polynomial Questions and Problems with Solutions

Polynomial questions and problems related to graphs, x and y intercepts, coefficients, degree, leading coefficients, ... with detailed solutions are presented.

6.5 Polynomial Equations

Learning objectives.

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

Use the Zero Product Property

Solve quadratic equations by factoring
Solve equations with polynomial functions
Solve applications modeled by polynomial equations

Be Prepared 6.10

Before you get started, take this readiness quiz.

Solve: 5 y − 3 = 0 . 5 y − 3 = 0 . If you missed this problem, review Example 2.2 .

Be Prepared 6.11

Factor completely: n 3 − 9 n 2 − 22 n . n 3 − 9 n 2 − 22 n . If you missed this problem, review Example 3.48 .

Be Prepared 6.12

If f ( x ) = 8 x − 16 , f ( x ) = 8 x − 16 , find f ( 3 ) f ( 3 ) and solve f ( x ) = 0 . f ( x ) = 0 . If you missed this problem, review Example 3.59 .

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one . Polynomial equations of degree one are linear equations are of the form a x + b = c . a x + b = c .

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get n 2 + n . n 2 + n .

The general form of a quadratic equation is a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , with a ≠ 0 . a ≠ 0 . (If a = 0 , a = 0 , then 0 · x 2 = 0 0 · x 2 = 0 and we are left with no quadratic term.)

Quadratic Equation

An equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 is called a quadratic equation.

To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

We will first solve some quadratic equations by using the Zero Product Property . The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Zero Product Property

If a · b = 0 , a · b = 0 , then either a = 0 a = 0 or b = 0 b = 0 or both.

We will now use the Zero Product Property, to solve a quadratic equation .

Example 6.44

How to solve a quadratic equation using the zero product property.

Solve: ( 5 n − 2 ) ( 6 n − 1 ) = 0 . ( 5 n − 2 ) ( 6 n − 1 ) = 0 .

Try It 6.87

Solve: ( 3 m − 2 ) ( 2 m + 1 ) = 0 . ( 3 m − 2 ) ( 2 m + 1 ) = 0 .

Try It 6.88

Solve: ( 4 p + 3 ) ( 4 p − 3 ) = 0 . ( 4 p + 3 ) ( 4 p − 3 ) = 0 .

Use the Zero Product Property.

Step 1. Set each factor equal to zero.
Step 2. Solve the linear equations.
Step 3. Check.

Solve Quadratic Equations by Factoring

The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we must be sure to start with the quadratic equation in standard form , a x 2 + b x + c = 0 . a x 2 + b x + c = 0 . Then we must factor the expression on the left.

Example 6.45

How to solve a quadratic equation by factoring.

Solve: 2 y 2 = 13 y + 45 . 2 y 2 = 13 y + 45 .

Try It 6.89

Solve: 3 c 2 = 10 c − 8 . 3 c 2 = 10 c − 8 .

Try It 6.90

Solve: 2 d 2 − 5 d = 3 . 2 d 2 − 5 d = 3 .

Solve a quadratic equation by factoring.

Step 1. Write the quadratic equation in standard form, a x 2 + b x + c = 0 . a x 2 + b x + c = 0 .
Step 2. Factor the quadratic expression.
Step 3. Use the Zero Product Property.
Step 4. Solve the linear equations.
Step 5. Check. Substitute each solution separately into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form .

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Example 6.46

Solve: 169 x 2 = 49 . 169 x 2 = 49 .

We leave the check up to you.

Try It 6.91

Solve: 25 p 2 = 49 . 25 p 2 = 49 .

Try It 6.92

Solve: 36 x 2 = 121 . 36 x 2 = 121 .

In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property , one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Example 6.47

Solve: ( 3 x − 8 ) ( x − 1 ) = 3 x . ( 3 x − 8 ) ( x − 1 ) = 3 x .

Try It 6.93

Solve: ( 2 m + 1 ) ( m + 3 ) = 1 2 m . ( 2 m + 1 ) ( m + 3 ) = 1 2 m .

Try It 6.94

Solve: ( k + 1 ) ( k − 1 ) = 8 . ( k + 1 ) ( k − 1 ) = 8 .

In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Example 6.48

Solve: 3 x 2 = 12 x + 63 . 3 x 2 = 12 x + 63 .

Try It 6.95

Solve: 18 a 2 − 30 = −33 a . 18 a 2 − 30 = −33 a .

Try It 6.96

Solve: 123 b = −6 − 60 b 2 . 123 b = −6 − 60 b 2 .

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.

Example 6.49

Solve: 9 m 3 + 100 m = 60 m 2 . 9 m 3 + 100 m = 60 m 2 .

Try It 6.97

Solve: 8 x 3 = 24 x 2 − 18 x . 8 x 3 = 24 x 2 − 18 x .

Try It 6.98

Solve: 16 y 2 = 32 y 3 + 2 y . 16 y 2 = 32 y 3 + 2 y .

Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

Example 6.50

For the function f ( x ) = x 2 + 2 x − 2 , f ( x ) = x 2 + 2 x − 2 ,

ⓐ find x when f ( x ) = 6 f ( x ) = 6 ⓑ find two points that lie on the graph of the function.

ⓑ Since f ( −4 ) = 6 f ( −4 ) = 6 and f ( 2 ) = 6 , f ( 2 ) = 6 , the points ( −4 , 6 ) ( −4 , 6 ) and ( 2 , 6 ) ( 2 , 6 ) lie on the graph of the function.

Try It 6.99

For the function f ( x ) = x 2 − 2 x − 8 , f ( x ) = x 2 − 2 x − 8 ,

ⓐ find x when f ( x ) = 7 f ( x ) = 7 ⓑ Find two points that lie on the graph of the function.

Try It 6.100

For the function f ( x ) = x 2 − 8 x + 3 , f ( x ) = x 2 − 8 x + 3 ,

ⓐ find x when f ( x ) = −4 f ( x ) = −4 ⓑ Find two points that lie on the graph of the function.

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function .

Zero of a Function

For any function f , if f ( x ) = 0 , f ( x ) = 0 , then x is a zero of the function .

When f ( x ) = 0 , f ( x ) = 0 , the point ( x , 0 ) ( x , 0 ) is a point on the graph. This point is an x -intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.

Example 6.51

For the function f ( x ) = 3 x 2 + 10 x − 8 , f ( x ) = 3 x 2 + 10 x − 8 , find

ⓐ the zeros of the function, ⓑ any x -intercepts of the graph of the function, ⓒ any y -intercepts of the graph of the function

ⓐ To find the zeros of the function, we need to find when the function value is 0.

ⓑ An x -intercept occurs when y = 0 . y = 0 . Since f ( −4 ) = 0 f ( −4 ) = 0 and f ( 2 3 ) = 0 , f ( 2 3 ) = 0 , the points ( −4 , 0 ) ( −4 , 0 ) and ( 2 3 , 0 ) ( 2 3 , 0 ) lie on the graph. These points are x -intercepts of the function. ⓒ A y -intercept occurs when x = 0 . x = 0 . To find the y -intercepts we need to find f ( 0 ) . f ( 0 ) .

Since f ( 0 ) = −8 , f ( 0 ) = −8 , the point ( 0 , −8 ) ( 0 , −8 ) lies on the graph. This point is the y -intercept of the function.

Try It 6.101

For the function f ( x ) = 2 x 2 − 7 x + 5 , f ( x ) = 2 x 2 − 7 x + 5 , find

ⓐ the zeros of the function, ⓑ any x -intercepts of the graph of the function, ⓒ any y -intercepts of the graph of the function.

Try It 6.102

For the function f ( x ) = 6 x 2 + 13 x − 15 , f ( x ) = 6 x 2 + 13 x − 15 , find

Solve Applications Modeled by Polynomial Equations

The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.

Use a problem solving strategy to solve word problems.

Step 1. Read the problem. Make sure all the words and ideas are understood.
Step 2. Identify what we are looking for.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
Step 5. Solve the equation using appropriate algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a polynomial equation.

Example 6.52

The product of two consecutive odd integers is 323. Find the integers.

Try It 6.103

The product of two consecutive odd integers is 255. Find the integers.

Try It 6.104

The product of two consecutive odd integers is 483 Find the integers.

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

Example 6.53

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

Try It 6.105

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

Try It 6.106

A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

In the next example, we will use the Pythagorean Theorem ( a 2 + b 2 = c 2 ) . ( a 2 + b 2 = c 2 ) . This formula gives the relation between the legs and the hypotenuse of a right triangle.

We will use this formula to in the next example.

Example 6.54

A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.

Try It 6.107

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

Try It 6.108

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

Example 6.55

Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function h ( t ) = −16 t 2 + 64 t + 80 h ( t ) = −16 t 2 + 64 t + 80 models the height, h , of the ball above the ground as a function of time, t. Find:

ⓐ the zeros of this function which tell us when the ball hits the ground, ⓑ when the ball will be 80 feet above the ground, ⓒ the height of the ball at t = 2 t = 2 seconds.

ⓐ The zeros of this function are found by solving h ( t ) = 0 . h ( t ) = 0 . This will tell us when the ball will hit the ground.

The result t = 5 t = 5 tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result t = −1 t = −1 is discarded.

ⓑ The ball will be 80 feet above the ground when h ( t ) = 80 . h ( t ) = 80 .

Try It 6.109

Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function h ( t ) = −16 t 2 + 48 t + 160 h ( t ) = −16 t 2 + 48 t + 160 models the height, h , of the rock above the ocean as a function of time, t . Find:

ⓐ the zeros of this function which tell us when the rock will hit the ocean, ⓑ when the rock will be 160 feet above the ocean, ⓒ the height of the rock at t = 1.5 t = 1.5 seconds.

Try It 6.110

Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 128 h ( t ) = −16 t 2 + 32 t + 128 models the height, h , of the penny above the ocean as a function of time, t . Find:

ⓐ the zeros of this function which is when the penny will hit the ocean, ⓑ when the penny will be 128 feet above the ocean, ⓒ the height the penny will be at t = 1 t = 1 seconds which is when the penny will be at its highest point.

Access this online resource for additional instruction and practice with quadratic equations.

Beginning Algebra & Solving Quadratics with the Zero Property

Section 6.5 Exercises

Practice makes perfect.

In the following exercises, solve.

( 3 a − 10 ) ( 2 a − 7 ) = 0 ( 3 a − 10 ) ( 2 a − 7 ) = 0

( 5 b + 1 ) ( 6 b + 1 ) = 0 ( 5 b + 1 ) ( 6 b + 1 ) = 0

6 m ( 12 m − 5 ) = 0 6 m ( 12 m − 5 ) = 0

2 x ( 6 x − 3 ) = 0 2 x ( 6 x − 3 ) = 0

( 2 x − 1 ) 2 = 0 ( 2 x − 1 ) 2 = 0

( 3 y + 5 ) 2 = 0 ( 3 y + 5 ) 2 = 0

5 a 2 − 26 a = 24 5 a 2 − 26 a = 24

4 b 2 + 7 b = −3 4 b 2 + 7 b = −3

4 m 2 = 17 m − 15 4 m 2 = 17 m − 15

n 2 = 5 n − 6 n 2 = 5 n − 6

7 a 2 + 14 a = 7 a 7 a 2 + 14 a = 7 a

12 b 2 − 15 b = −9 b 12 b 2 − 15 b = −9 b

49 m 2 = 144 49 m 2 = 144

625 = x 2 625 = x 2

16 y 2 = 81 16 y 2 = 81

64 p 2 = 225 64 p 2 = 225

121 n 2 = 36 121 n 2 = 36

100 y 2 = 9 100 y 2 = 9

( x + 6 ) ( x − 3 ) = −8 ( x + 6 ) ( x − 3 ) = −8

( p − 5 ) ( p + 3 ) = −7 ( p − 5 ) ( p + 3 ) = −7

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

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

( 3 x − 2 ) ( x + 4 ) = 12 x ( 3 x − 2 ) ( x + 4 ) = 12 x

( 2 y − 3 ) ( 3 y − 1 ) = 8 y ( 2 y − 3 ) ( 3 y − 1 ) = 8 y

20 x 2 − 60 x = −45 20 x 2 − 60 x = −45

3 y 2 − 18 y = −27 3 y 2 − 18 y = −27

15 x 2 − 10 x = 40 15 x 2 − 10 x = 40

14 y 2 − 77 y = −35 14 y 2 − 77 y = −35

18 x 2 − 9 = −21 x 18 x 2 − 9 = −21 x

16 y 2 + 12 = −32 y 16 y 2 + 12 = −32 y

16 p 3 = 24 p 2 – 9 p 16 p 3 = 24 p 2 – 9 p

m 3 − 2 m 2 = − m m 3 − 2 m 2 = − m

2 x 3 + 72 x = 24 x 2 2 x 3 + 72 x = 24 x 2

3 y 3 + 48 y = 24 y 2 3 y 3 + 48 y = 24 y 2

36 x 3 + 24 x 2 = −4 x 36 x 3 + 24 x 2 = −4 x

2 y 3 + 2 y 2 = 12 y 2 y 3 + 2 y 2 = 12 y

For the function, f ( x ) = x 2 − 8 x + 8 , f ( x ) = x 2 − 8 x + 8 , ⓐ find when f ( x ) = −4 f ( x ) = −4 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = x 2 + 11 x + 20 , f ( x ) = x 2 + 11 x + 20 , ⓐ find when f ( x ) = −8 f ( x ) = −8 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = 8 x 2 − 18 x + 5 , f ( x ) = 8 x 2 − 18 x + 5 , ⓐ find when f ( x ) = −4 f ( x ) = −4 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = 18 x 2 + 15 x − 10 , f ( x ) = 18 x 2 + 15 x − 10 , ⓐ find when f ( x ) = 15 f ( x ) = 15 ⓑ Use this information to find two points that lie on the graph of the function.

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

f ( x ) = 9 x 2 − 4 f ( x ) = 9 x 2 − 4

f ( x ) = 25 x 2 − 49 f ( x ) = 25 x 2 − 49

f ( x ) = 6 x 2 − 7 x − 5 f ( x ) = 6 x 2 − 7 x − 5

f ( x ) = 12 x 2 − 11 x + 2 f ( x ) = 12 x 2 − 11 x + 2

Solve Applications Modeled by Quadratic Equations

The product of two consecutive odd integers is 143. Find the integers.

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

The product of two consecutive even integers is 288. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

A rectangular carport has area 150 square feet. The width of the carport is five feet less than twice its length. Find the width and the length of the carport.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure.

Juli is going to launch a model rocket in her back yard. When she launches the rocket, the function h ( t ) = −16 t 2 + 32 t h ( t ) = −16 t 2 + 32 t models the height, h , of the rocket above the ground as a function of time, t . Find:

ⓐ the zeros of this function, which tell us when the rocket will be on the ground. ⓑ the time the rocket will be 16 feet above the ground.

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 48 h ( t ) = −16 t 2 + 32 t + 48 models the height, h , of the ball above the ground as a function of time, t . Find:

ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 48 feet above the ground. ⓒ the height the ball will be at t = 1 t = 1 seconds which is when the ball will be at its highest point.

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

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

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

Authors: Lynn Marecek, Andrea Honeycutt Mathis
Publisher/website: OpenStax
Book title: Intermediate Algebra 2e
Publication date: May 6, 2020
Location: Houston, Texas
Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/6-5-polynomial-equations

© Jan 23, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Mathematicians
Math Lessons
Square Roots
Math Calculators
Polynomial equation – Properties, Techniques, and Examples

JUMP TO TOPIC

Polynomial equation definition and examples

Properties and techniques in solving polynomial equations, finding the zeros of a polynomial function, practice questions, polynomial equation – properties, techniques, and examples.

Polynomial equations are equations that contain polynomials on both sides of the equation.

Since we’re dealing with polynomials and polynomial functions, make sure to check out our article on polynomial functions .

Polynomial equations such as quadratic functions are often used in modeling motions, real-world functions, and extensive technology and science applications. This is also why we need to understand how we can identify and solve polynomial equations .

What is a polynomial equation?

Polynomial equations are equations that contain polynomial expressions on both sides of the equation. Here’s the standard form of a polynomial equation.

Note that a n , a n-1 , … a o can be any complex number, and the exponents can only be whole numbers for these to be considered polynomial expressions.

Having an equal sign followed by another polynomial expression makes the polynomial equation distinct from polynomial expressions.

As can be confirmed from the equation shown above, the polynomial equation is said to be in standard form when the terms are arranged from the term with the highest power to the one with the lowest power.

Polynomial equations contain polynomial expressions, so properties of polynomial functions will still apply. In fact, the degree and the number of terms of the polynomial expression can also help us classify polynomial equations.

Let’s go ahead and take a look at the common types of polynomial equations we may encounter based on the degree:

How to solve polynomial equations?

A great way to visualize polynomial equations is to think of it is as the result of combining different blocks. When the goal is to find the roots, solutions, or solving for the polynomial equation, we must find a way to take each block apart.

Here are some important pointers to remember when solving polynomial equations:

If the polynomial equation is still not in its standard form, rewrite the equation so that it is in standard form : all polynomial expressions on the left side and 0 on the right.
Simplify the polynomial equation in standard form and predict the number of zeroes or roots that the equation might have.
If the polynomial equation is a linear or quadratic equation, apply previous knowledge to solve these types of equations .
If the polynomial equation has a three or higher degree, start by finding one rational factor or zero .
Take out this factor and repeat the same process until you’re left with a linear equation or a constant .
List down all the roots or zeros.

Let’s do a quick recap of the different techniques we can apply to achieve Step 3. As mentioned, at this point, we should know how to solve linear and quadratic equations extensively. Don’t worry. This website contains a handful of resources about these two equations in case we need a quick refresher.

Linear Equations

Linear equations are polynomial equations that have a degree of 1.

Solving for solutions for this type of equation will require us to isolate the unknown variable on one side of the equation. Master your craft in solving linear equations here .

Quadratic Equations

Quadratic equations are polynomial equations with a degree of 2.

ax 2 + bx + c = 0

There are different ways we can solve quadratic equations – it mostly depends on the form of the quadratic expression on the right-hand side.

We can factor quadratic expressions and apply the zero-property.
Applying special algebraic properties such as the difference of two squares , perfect square trinomial properties, and completing the square .
Lastly, we can also use the quadratic formula to find the zeroes of quadratic equations.

Polynomial Equations (with a degree of 3 or higher)

Here’s the exciting part: what if we need to find the zeros of the solutions of a polynomial equation with degrees that are 3 or higher?

Some cubic and quartic equations can be factored by grouping and be reduced to equations with a smaller degree. There are times, however, that finding the actual factors can be challenging.

Here are important properties of polynomial equations that we’ll need to understand to easily find the real zeros and roots of a polynomial equation.

Real Zeros in a Function

The num ber of real zeroes a polynomial function can have is the same value of the degree. What does this mean? If f(x) has a degree of 5, the maximum number of real zeroes it can have is 5.

Descartes’ Rule of Signs

This rule is helpful when we need to find the zeroes of a polynomial equation without its graph. What does Descartes’s Rule of Signs do? It tells us the number and position of a polynomial equation’s zeroes.

To apply this rule, we’ll need to observe the signs between the coefficients of both f(x) and f(-x). Let’s say we have f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12.

Count the number of times the coefficients switch signs, and the table below summarizes what the result means:

Let’s apply this with f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12.

f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12 f(-x) = 2x 4 + 2x 3 – 14x 2 – 2x + 12

From the sign changes in f(x), there can be 2 or 0 positive real zeros. Similarly, from f(-x), there are can also be 2 or 0 negative real zeros.

Rational Zeros Theorem

This theorem will help us narrow down the possible rational zeros of a polynomial function . Let p contain all the factors of a n (leading term) and q contain all the factors from a o (constant term).

The possible rational zeros of the polynomial equation can be from dividing p by q, p/q . Make sure that the list contains all possible expressions for p/q in the lowest form.

Using the same example, f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12, we have p = 2 and q = 12 . Let’s go ahead and list down all the possible rational zeros of f(x).

Does this mean f(x) has 14 rational zeros? No, this list tells us that if f(x) has rational zeros, it will come from this list. Meaning , we have reduced the possibilities to a reasonable number from an extensive range of rational numbers .

Applying the Remainder Theorem and Synthetic Division

How do we slowly find rational zeros of f(x) once we have a list of p/q? It’s time that we apply our past knowledge on the remainder theorem , factor theorem , and synthetic division . Make sure to take a quick refresher for these topics by clicking on the links.

This means that we can try each factor listed and check for the remainder.
If the remainder is zero, the value of p/q is a root of f(x).
Use the resulting polynomials and repeat the same process until we have all the f(x) zeros.

Why don’t we apply what we’ve just learned to find the zeros of 2x 4 – 2x 3 – 14x 2 + 2x + 12 = 0? From the previous section, we’ve seen that f(x)’s list of possible rational zeroes is: ±1/12, ±1/6, ±1/4, ± 1/3, ±1/2, ±2/3, ±1, and ±2.

Let’s check if x = 1 is a root of f(x) using synthetic division.

1 | 2 -2 -14 2 12

2 0 -14 -12

_____________________________

2 0 -14 -12 0

Since the remainder is 0, (x – 1) is a factor of f(x) and x = 1 a solution to the equation . Let’s express f(x) as a factor of (x – 1): f(x) = (x -1)(2x 3 – 14x – 12).

Use the resulting cubic expression and find a second root for the equation. Let’s try to see if x = 2 is a root of 2x 3 – 14x – 12.

-2 | 2 0 -14 -12

-4 8 12

__________________________

2 -4 -6 0

Hence, (x + 2) is a factor of f(x) and x = -2 is a root of the equation . Since we have a quadratic expression, we can factor the expression and solve for the two remaining zeros of the equation.

2x 2 – 4x – 6 = 0

2(x 2 – 2x – 3) = 0

x 2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x = 3, x = -1

Let’s go ahead and take note of all the zeros we have solved for our polynomial equation: x = 1, x = -2, x = 3, and x = -1 . The equation has four zeros, as we have expected since it has a degree of 4.

Apply a similar process when finding the zeros of other polynomial equations.

Here’s a guide to help you summarize and follow through the steps we might need to do when finding the zeroes of a given polynomial equation in standard form:

The guide above helps you ask the right questions to ensure we apply the best strategy when solving polynomial equations.

Why don’t we apply this by answering the questions shown below?

Given that f(x) = -2x 3 + 4x 2 – 7x – 6, how many sign changes are there in f(x) and f(-x)? Interpret the results.

We can immediately inspect f(x) for its sign changes. We have two sign changes : -2x 3 and 4x 2 and as +4x 2 and -7x.

As for f(-x), let’s go ahead and find the expression for f(-x) first.

f(-x) = 2x 3 + 4x 2 + 7x -6

From this, we can see that f(-x) has only one sign change : between 7x and -6. Using Descartes’ Rule of Sign, we can conclude that:

When f(x) is equated to 0, the resulting equation may have two or zero positive real zeros .
Similarly, the resulting equation may have one negative real zero .

True or False? Given that the polynomial function, g(x), has a degree of 3, the equation g(x) = 0 will always have three real zeros.

The equation g(x) = 0 will have at most three possible real zeros. This means that it may or may not have three real zeros exactly. One function to further show that the statement is not true is when g(x) = x 3 + x.

Let’s solve the equation and observe the results for x. Since the expression is still factorable, we’ll factor x out and equate x 2 + 1 to 0.

x 3 + x = 0

x(x 2 + 1) = 0

x 2 + 1 = 0

This will only be true when x = -i or x = i. This clearly shows that it is possible for g(x) to not have three real zeros despite having a degree of 3. Hence, the statement is not true.

Find the values of x that satisfies the given equation: 4x 5 – 4x 4 + 73x 2 = -18(x -1)+ 73x 3 .

The equation is still not in its standard form, so let’s go ahead and isolate all terms on the left-hand side.

4x 5 – 4x 4 – 73x 3 + 73x 2 + 18x – 18 = 0

Using the rational zeros theorem, let’s list down the possible rational zeros for the polynomial equation.

1 | 4 -4 -73 73 18 -18

4 0 -73 0 18

____________________________________

4 0 -73 0 18 0

Since the remainder is 0, (x – 1) is a factor of the expression and x = 1 is a solution. Let’s go ahead and try x = 1/2 and x = -1/2 to see if they are zeros of the equation too.

1/2 | 4 0 -73 0 18

2 1 -36 -18

_________________________________

4 2 -72 -36 0

Make sure to check with resulting polynomial. We now have (x – 1)(x – 1/2)(4x 3 + 2x 2 – 72x – 36) = 0.

-1/2 | 4 2 -72 -36

-2 0 36

4 0 -72 0

We can see that x = -1/2 and x = 1/2 are both zeros of the polynomial equation from these two consecutive synthetic divisions.

We now have (x – 1)(x – 1/2)(x + 1/2)(4x 2 – 72) = 0. Since the remaining expression is a quadratic expression, we can equate it to 0 and solve the polynomial equation’s remaining zeros.

4x 2 – 72 = 0

4(x 2 – 18) = 0

x 2 – 18 = 0

We now have five zeros for the polynomial equation (this is already the maximum number of zeros possible for a polynomial equation with a degree of 5).

Hence, the equation has a solution set of: {-2√2, -1/2, 1/2, 1, 2√2}.

Previous Lesson | Main Page | Next Lesson

Module 7: Polynomials and Polynomial Functions

7.5 – polynomial equations and applications, learning objectives.

The principle of zero products
Projectile motion
Using the Pythagorean theorem to find the lengths of a right triangle

Geometric Applications

Cost, revenue, and profit polynomials

(7.5.1) – Solving polynomial equations

Not all of the techniques we use for solving linear equations will apply to solving polynomial equations. In this section we will introduce a method for solving polynomial equations that combines factoring and the zero product principle.

The Principle of Zero Products

What if we told you that we multiplied two numbers together and got an answer of zero? What could you say about the two numbers? Could they be 2 and 5? Could they be 9 and 1? No! When the result (answer) from multiplying two numbers is zero, that means that one of them had to be zero. This idea is called the zero product principle, and it is useful for solving polynomial equations that can be factored.

Principle of Zero Products

The Principle of Zero Products states that if the product of two numbers is 0, then at least one of the factors is 0. If [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or both [latex]a[/latex] and [latex]b[/latex] are 0.

Let’s start with a simple example. We will factor a GCF from a binomial and apply the principle of zero products to solve a polynomial equation.

[latex]-t^2+t=0[/latex]

Each term has a common factor of [latex]t[/latex], so we can factor and use the zero product principle. Rewrite each term as the product of the GCF and the remaining terms.

[latex]\begin{array}{c}-t^2=t\left(-t\right)\\t=t\left(1\right)\end{array}[/latex]

Rewrite the polynomial equation using the factored terms in place of the original terms.

[latex]\begin{array}{c}-t^2+t=0\\t\left(-t\right)+t\left(1\right)\\t\left(-t+1\right)=0\end{array}[/latex]

Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.

[latex]\begin{array}{c}t=0\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,-t+1=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-1}\,\,\,\underline{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-t=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-t}{-1}=\frac{-1}{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=1\end{array}[/latex]

[latex]t=0\text{ OR }t=1[/latex]

In the following video we show two more examples of using both factoring and the principle of zero products to solve a polynomial equation.

In the next video we show that you can factor a trinomial using methods previously learned to solve a quadratic equation.

You and I both know that it is rare to be given an equation to solve that has zero on one side, so let’s try another one.

Solve: [latex]s^2-4s=5[/latex]

First, move all the terms to one side. The goal is to try and see if we can use the zero product principle, since that is the only tool we know for solving polynomial equations.

[latex]\begin{array}{c}\,\,\,\,\,\,\,s^2-4s=5\\\,\,\,\,\,\,\,s^2-4s-5=0\\\end{array}[/latex]

We now have all the terms on the left side, and zero on the other side. The polynomial [latex]s^2-4s-5[/latex] factors nicely, which makes this equation a good candidate for the zero product principle. (imagine that)

[latex]\begin{array}{c}s^2-4s-5=0\\\left(s+1\right)\left(s-5\right)=0\end{array}[/latex]

We separate our factors into two linear equations using the principle of zero products.

[latex]\begin{array}{c}\left(s-5\right)=0\\s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array}[/latex]

[latex]\begin{array}{c}\left(s+1\right)=0\\s+1=0\\s=-1\end{array}[/latex]

[latex]s=-1\text{ OR }s=5[/latex]

We will work through one more example that is similar to the one above, except this example has fractions.

Solve [latex]y^2-5=-\frac{7}{2}y+\frac{5}{2}[/latex]

Start by multiplying the whole equation by 2 to eliminate fractions:

[latex]\begin{array}{ccc}2\left(y^2-5=-\frac{7}{2}y+\frac{5}{2}\right)\\\,\,\,\,\,\,2(y^2)+2(-5)=2\left(-\frac{7}{2}\right)+2\left(-\frac{5}{2}\right)\\2y^2-10=-7y+5\end{array}[/latex]

Now we can move all the terms to one side and see if this will factor so we can use the principle of zero products.

[latex]\begin{array}{c}2y^2-10=-7y-5\\2y^2-10+7y-5=0\\2y^2-15+7y=0\\2y^2+7y-15=0\end{array}[/latex]

We can now check whether this polynomial will factor, using a table we can list factors until we find two numbers with a product of [latex]2\cdot-15=-30[/latex] and a sum of 7.

We have found the factors that will produce the middle term we want,[latex]-3,10[/latex]. We need to place the factors in a way that will lead to a term of [latex]10y[/latex]:

[latex]\left(2y-3\right)\left(y+5\right)=0[/latex]

Now we can set each factor equal to zero and solve:

[latex]\begin{array}{ccc}\left(2y-3\right)=0\text{ OR }\left(y+5\right)=0\\2y=3\text{ OR }y=-5\\y=\frac{3}{2}\text{ OR }y=-5\end{array}[/latex]

You can always check your work to make sure your solutions are correct:

Check [latex]y=\frac{3}{2}[/latex]

[latex]\begin{array}{ccc}\left(\frac{3}{2}\right)^2-5=-\frac{7}{2}\left(\frac{3}{2}\right)+\frac{5}{2}\\\frac{9}{4}-5=-\frac{21}{4}+\frac{5}{2}\\\text{ common denominator = 4}\\\frac{9}{4}-\frac{20}{4}=-\frac{21}{4}+\frac{10}{4}\\-\frac{11}{4}=-\frac{11}{4}\end{array}[/latex]

[latex]y=\frac{3}{2}[/latex] is indeed a solution, now check [latex]y=-5[/latex]

[latex]\begin{array}{ccc}\left(-5\right)^2-5=-\frac{7}{2}\left(-5\right)+\frac{5}{2}\\25-5=\frac{35}{2}+\frac{5}{2}\\20=\frac{40}{2}\\20=20\end{array}[/latex]

[latex]y=-5[/latex] is also a solution, so we must have done something right!

[latex]y=\frac{3}{2}\text{ OR }y=-5[/latex]

In our last video, we show how to solve another quadratic equation that contains fractions.

(7.5.2) – Applications of polynomial equations

Projectile motion.

Projectile motion happens when you throw a ball into the air and it comes back down because of gravity. A projectile will follow a curved path that behaves in a predictable way. This predictable motion has been studied for centuries, and in simple cases it’s height from the ground at a given time, [latex]t[/latex], can be modeled with a polynomial function of the form [latex]h(t)=at^2+bt+c[/latex], where h(t) = height of an object at a given time, [latex]t[/latex]. Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile’s motion, as in the image of water in the fountain below.

Water from a fountain shoing classic parabolic motion.

Parabolic WaterTrajectory

Parabolic motion and it’s related functions allow us to launch satellites for telecommunications, and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles, rather than pursuing them by high-speed chase [1] .

In this section we will use polynomial functions to answer questions about the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial function that models projectile motion.

A small toy rocket is launched from a 4-foot pedestal. The height ([latex]h[/latex] , in feet) of the rocket [latex]t[/latex] seconds after taking off is given by the function [latex]h(t)=−2t^{2}+7t+4[/latex]. How long will it take the rocket to hit the ground?

The rocket will be on the ground when the [latex]h(t)=0[/latex]. We want to know how long, [latex]t[/latex], the rocket is in the air.

[latex]\begin{array}{l}h(t)=−2t^{2}+7t+4=0\\0=−2t^{2}+7t+4\end{array}[/latex]

We can factor the polynomial [latex]−2t^{2}+7t+4[/latex] more easily by first factoring out a [latex]-1[/latex]

[latex]\begin{array}{c}0=-1(2t^{2}-7t-4)\\0=-1\left(2t+1\right)\left(t-4\right)\end{array}[/latex]

Use the Zero Product Property. There is no need to set the constant factor [latex]-1[/latex] to zero, because [latex]-1[/latex] will never equal zero.

[latex]2t+1=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t-4=0[/latex]

Solve each equation.

[latex]t=-\frac{1}{2}\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t=4[/latex]

Interpret the answer. Since t represents time, it cannot be a negative number; only [latex]t=4[/latex] makes sense in this context.

[latex]t=4[/latex]

We can check our answer: [latex]\begin{array}{c}h(4)=−2(4)^{2}+7(4)+4=0\\h(4)=-2(16)+28+4=0\\h(4)-32+32=0\\h(4)=0\end{array}[/latex]

The rocket will hit the ground 4 seconds after being launched.

In the next example we will solve for the time that the rocket is at a given height other than zero.

Use the formula for the height of the rocket in the previous example to find the time when the rocket is 4 feet from hitting the ground on it’s way back down. Refer to the image.

[latex]h(t)=−2t^{2}+7t+4[/latex]

Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.

We are given that the height of the rocket is 4 feet from the ground on it’s way back down. We want to know how long it has taken the rocket to get to that point in it’s path, we are going to solve for [latex]t[/latex].

Substitute [latex]h(t) = 4[/latex] into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials.

Write and Solve:

[latex]\begin{array}{l}h(t)=4=−2t^{2}+7t+4\\4=-2t^2+7t+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\0=-2t^2+7t\end{array}[/latex]

Now we can factor out a [latex]t[/latex] from each term:

[latex]0=t\left(-2t+7\right)[/latex]

Solve each equation for [latex]t[/latex] using the zero product principle:

[latex]\begin{array}{l}t=0\text{ OR }-2t+7=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-7}\,\,\,\,\,\,\,\underline{-7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-2t}{-2}=\frac{-7}{-2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=\frac{7}{2}=3.5\end{array}[/latex]

It doesn’t make sense for us to choose [latex]t=0[/latex] because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it’s way back down. We will choose [latex]t=3.5[/latex]

Check the answer on your own for practice.

[latex]t=3.5\text{ seconds }[/latex]

In the following video we show another example of how to find the time when a object following a parabolic trajectory hits the ground.

In this section we introduced the concept of projectile motion, and showed that it can be modeled with polynomial function. While the models used in these examples are simple, the concepts and interpretations are the same as what would happen in “real life”.

Pythagorean Theorem

The Pythagorean theorem or Pythagoras’s theorem is a statement about the sides of a right triangle. One of the angles of a right triangle is always equal to 90 degrees. This angle is the right angle. The two sides next to the right angle are called the legs and the other side is called the hypotenuse. The hypotenuse is the side opposite to the right angle, and it is always the longest side. The image above shows four common kinds of triangle, including a right triangle.

Right Triangle Labeled

The Pythagorean theorem is often used to find unknown lengths of the sides of right triangles. If the longest leg of a right triangle is labeled c, and the other two a, and b as in the image on teh left, The Pythagorean Theorem states that

[latex]a^2+b^2=c^2[/latex]

Given enough information, we can solve for an unknown length. This relationship has been used for many, many years for things such as celestial navigation and early civil engineering projects. We now have digital GPS and survey equipment that have been programmed to do the calculations for us.

In the next example we will combine the power of the Pythagorean theorem and what we know about solving quadratic equations to find unknown lengths of right triangles.

A right triangle has one leg with length [latex]x[/latex], another whose length is greater by two, and the length of the hypotenuse is greater by four. Find the lengths of the sides of the triangle. Use the image below.

Right triangle with one leg having length = x, one with length= x+2 and the hypotenuse = x+4

Read and understand: We know the lengths of all the sides of a triangle in terms of one side. We also know that the Pythagorean theorem will give us a relationship between the side lengths of a right triangle.

Translate:

[latex]\begin{array}{l}a^2+b^2=c^2\\x^2+\left(x+2\right)^2=\left(x+4\right)^2\end{array}[/latex]

Solve: If we can move all the terms to one side and factor, we can use the zero product principle to solve. Since this is the only method we know – let’s hope it works!

First, multiply the binomials and simplify so we can see what we are working with.

[latex]\begin{array}{l}x^2+\left(x+2\right)^2=\left(x+4\right)^2\\x^2+x^2+4x+4=x^2+8x+16\\2x^2+4x+4=x^2+8x+16\end{array}[/latex]

Now move all the terms to one side and see if we can factor.

[latex]\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\,\,\,\,\,\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\\x^2-4x-12=0\end{array}[/latex]

This went from a messy looking problem to something promising. We can factor using the shortcut:

[latex]-6\cdot{2}=-12,\text{ and }-6+2=-4[/latex]

So we can build our binomial factors with -6 and 2:

[latex]\left(x-6\right)\left(x+2\right)=0[/latex]

Set each factor equal to zero:

[latex]x-6=0, x=6[/latex]

[latex]x+2=0, x=-2[/latex]

Interpret: Ok, it doesn’t make sense to have a length equal to -2, so we can safely throw that solution out. The lengths of the sides are as follows:

[latex]x=6[/latex]

[latex]x+2=6+2=8[/latex]

[latex]x+4=6+4=10[/latex]

Check: Since we know the relationship between the sides of a right triangle we can check that we are correct. Sometimes it helps to draw a picture

We know that [latex]a^2+b^2=c^2[/latex], so we can substitute the values we found:

[latex]\begin{array}{l}6^2+8^2=10^2\\36+64=100\\100=100\end{array}[/latex]

Our solution checks out.

The lengths of the sides of the right triangle are 6, 8, and 10

This video example shows another way a quadratic equation can be used to find and unknown length of a right triangle.

In this section we will explore ways that polynomials are used in applications of area problems.

The length of a rectangle is 3 more than the width. If the area is 40 square inches, what are the dimensions?

[latex]\begin{eqnarray*} x & & \text{ We} \text{ do} \text{ not} \text{ know} \text{ the} \text{ width}, x.\\ x + 3 & & \text{ Length} \text{ is}\, 3 \text{ more}, \text{ or}\, x + 3, \text{ and} \text{ area} \text{ is}\, 40.\\ x (x + 3) = 40 & & \text{ Multiply} \text{ length} \text{ by} \text{ width} \text{ to} \text{ get} \text{ area}\\ x^2 + 3 x = 40 & & \text{ Distribute}\\ \underline{- 40 - 40} & & \text{ Make} \text{ equation} \text{ equal} \text{ zero}\\ x^2 + 3 x - 40 = 0 & & \text{ Factor}\\ (x - 5) (x + 8) = 0 & & \text{ Set} \text{ each} \text{ factor} \text{ equal} \text{ to} \text{ zero}\\ x - 5 = 0 \text{ or}\, x + 8 = 0 & & \text{ Solve} \text{ each} \text{ equation}\\ x = 5 \text{ or}\, x = - 8 & & \text{ Our}\, x \text{ is} a \text{ width}, \text{ cannot} \text{ be} \text{ negative} .\\ (5) + 3 = 8 & & \text{ Length} \text{ is}\, x + 3, \text{ substitute} 5 \text{ for}\, x \text{ to} \text{ find} \text{ length}\\ 5 \text{ in} \text{ by}\, 8 \text{ in} & & \text{ Our} \text{ Solution} \end{eqnarray*}[/latex]

The dimensions of the rectangle are 5 inches by 8 inches.

In the following video you are shown how to find the dimensions of a border around a rectangle.

Cost, Revenue, and Profit Polynomials

In the systems of linear equations section, we discussed how a company’s cost and revenue can be modeled with two linear equations. We found that the profit region for a company was the area between the two lines where the company would make money based on how much was produced. In this section, we will see that sometimes polynomials are used to describe cost and revenue.

Profit is typically defined in business as the difference between the amount of money earned (revenue) by producing a certain number of items and the amount of money it takes to produce that number of items. When you are in business, you definitely want to see profit, so it is important to know what your cost and revenue is.

Cell Phones

For example, let’s say that the cost to a manufacturer to produce a certain number of things is C and the revenue generated by selling those things is R. The profit, P, can then be defined as

The example we will work with is a hypothetical cell phone manufacturer whose cost to manufacture x number of phones is [latex]C=2000x+750,000[/latex], and the Revenue generated from manufacturing x number of cell phones is [latex]R=-0.09x^2+7000x[/latex].

Define a Profit polynomial for the hypothetical cell phone manufacturer.

Read and Understand: Profit is the difference between revenue and cost, so we will need to define P = R – C for the company.

Define and Translate: [latex]\begin{array}{c}R=-0.09x^2+7000x\\C=2000x+750,000\end{array}[/latex]

Write and Solve: Substitute the expressions for R and C into the Profit equation.

[latex]\begin{array}{c}P=R-C\\=-0.09x^2+7000x-\left(2000x+750,000\right)\\=-0.09x^2+7000x-2000x-750,000\\=-0.09x^2+5000x-750,000\end{array}[/latex]

Remember that when you subtract a polynomial, you have to subtract every term of the polynomial.

[latex]P=-0.09x^2+5000x-750,000[/latex]

Mathematical models are great when you use them to learn important information. The cell phone manufacturing company can use the profit equation to find out how much profit they will make given [latex]x[/latex] number of phones are manufactured. In the next example, we will explore some profit values for this company.

Given the following numbers of cell phones manufactured, find the profit for the cell phone manufacturer:

x = 100 phones
x = 25,000 phones
x=60,000 phones

Interpret your results.

Read and Understand: The profit polynomial defined in the previous example,[latex]P=-0.09x^2+5000x-750,000[/latex], gives profit based on x number of phones manufactured. We need to substitute the given numbers of phones manufactured into the equation, then try to understand what our answer means in terms of profit and number of phones manufactured.

We will move straight into write and solve since we already have our polynomial. It is probably easiest to use a calculator since the numbers in this problem are so large.

Write and Solve:

Substitute [latex]x = 100[/latex]

[latex]\begin{array}{c}P=-0.09x^2+5000x-750,000\\=-0.09\left(100\right)^2+5000\left(100\right)-750,000\\=-900+500,000-750,000\\=-250,900\end{array}[/latex]

Interpret: When the number of phones manufactured is 100, the profit for the business is $-250,000. This is not what we want! The company must produce more than 100 phones to make a profit.

Substitute [latex]x = 25,000[/latex]

[latex]\begin{array}{c}P=-0.09x^2+5000x-750,000\\=-0.09\left(25000\right)^2+5000\left(25000\right)-750,000\\=-6120000+125,000,000-750,000\\=118,130,000\end{array}[/latex]

Interpret: When the number of phones manufactured is 25,000, the profit for the business is $118,130,000. This is more like it! If the company makes 25,000 phones it will make a profit after it pays all it’s bills.

If this is true, then the company should make even more phones so it can make even more money, right? Actually, something different happens as the number of items manufactured increases without bound.

Substitute [latex]x = 60,000[/latex]

[latex]\begin{array}{c}P=-0.09x^2+5000x-750,000\\=-0.09\left(60000\right)^2+5000\left(60000\right)-750,000\\=-324,000,000+300,000,000-750,000\\=-24,750,000\end{array}[/latex]

Interpret: When the number of phones manufactured is 60,000, the profit for the business is $-24,750,000. Wait a minute! If the company makes 60,000 phones it will lose money, what happened? At some point, the cost to manufacture the phones will overcome the amount of profit that the business can make. If this is interesting to you, you may enjoy reading about Economics and Business models.

In the video that follows, we present another example of finding a polynomial profit equation.

"Cops' Latest Tool in High-speed Chases: GPS Projectiles." CBSNews . CBS Interactive, n.d. Web. 14 June 2016. ↵
Solve a Quadratic Equations with Fractions by Factoring (a not 1). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/kDj_qdKW-ls . License : CC BY: Attribution
Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
Parabolic motion description and example. Provided by : Lumen Learning. License : CC BY: Attribution
Factoring Application - Find the Time When a Projectile Hits and Ground. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/hsWSzu3KcPU . License : CC BY: Attribution
Pythagorean Theorem, Description and Examples. Provided by : Lumen Learning. License : CC BY: Attribution
Screenshot: Thumbsdown. Provided by : Lumen Learning. License : CC BY: Attribution
Screenshot PI. Provided by : Lumen Learning. License : CC BY: Attribution
Polynomial Multiplication Application - Volume of a Cylinder. . Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/g-g_nSsfGs4 . License : CC BY: Attribution
Polynomial Subtracton App - Profit Equation from Revene and Cost.. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning.. Located at : https://youtu.be/-TWjDC4g9dU . License : CC BY: Attribution
Screenshot: Cell Phones.. Authored by : Lumen Learning. License : CC BY: Attribution
Profit Polynomial Examples. Provided by : Lumen Learning. License : CC BY: Attribution
Ex: Factor and Solve Quadratic Equation - Greatest Common Factor Only. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/gIwMkTAclw8 . License : CC BY: Attribution
Ex: Factor and Solve Quadratic Equations When A equals 1. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/bi7i_RuIGl0 . License : CC BY: Attribution
Unit 12: Factoring, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology. Located at : http://nrocnetwork.org/dm-opentext . License : CC BY: Attribution
Parabolic water trajectory. Authored by : By GuidoB. Located at : https://commons.wikimedia.org/w/index.php?curid=8015696 . License : CC BY-SA: Attribution-ShareAlike
Pythagorean Theorem. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/Pythagorean_theorem . License : CC BY-SA: Attribution-ShareAlike
Ex: Polynomial Addition Application - Perimeter.. Authored by : James Sousa (Mathispower4u.com). . Located at : . License : CC BY: Attribution
Ex: Find the Area of a Rectangle Using a Polynomial.. Authored by : James Sousa (Mathispower4u.com) .. Located at : . License : CC BY: Attribution
Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. . Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/dm-opentext . License : CC BY: Attribution
Applied Optimization Problems. Provided by : OpenStax. Located at : http://cnx.org/contents/svyieFe9@2/Applied-Optimization-Problems . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike . License Terms : Download for free at http://cnx.org/contents/b2fca278-57bd-421d-aa85-21f539b4cc6f@2
Applications of Quadratic Equations. Authored by : James Sousa. Located at : http://www.youtube.com/watch?v=AlIoxXQ-V50 . License : CC BY: Attribution
Quadratics - Rectangle. Authored by : Tyler Wallace. Located at : http://wallace.ccfaculty.org/book/book.html . Project : Beginning and Intermediate Algebra. License : Public Domain: No Known Copyright
Question ID#74417. Authored by : Nearing,Daniel. License : CC BY: Attribution
Celestial Navigation Math. Authored by : TabletClass Math. Located at : https://www.youtube.com/watch?v=XWLZKmPU17M . License : All Rights Reserved . License Terms : Standard YouTube License

PRACTICE PROBLEMS ON SOLVING POLYNOMIAL EQUATIONS

(1) Solve the cubic equation : 2x 3 − x 2 −18x + 9 = 0, if sum of two of its roots vanishes Solution

(2) Solve the equation 9x 3 − 36x 2 + 44x −16 = 0 if the roots form an arithmetic progression. Solution

(3) Solve the equation 3x 3 − 26x 2 + 52x − 24 = 0 if its roots form a geometric progression. Solution

(4) Determine k and solve the equation 2x 3 − 6x 2 + 3x + k = 0 if one of its roots is twice the sum of the other two roots. Solution

(5) Find all zeros of the polynomial x 6 − 3x 5 − 5x 4 + 22x 3 − 39x 2 − 39x + 135, if it is known that 1 + 2i an d √ 3 are two of its zeros. Solution

(6) Solve the cubic equation

(i) 2x 3 − 9x 2 +10x = 3

(ii) 8x 3 − 2x 2 − 7x + 3 = 0. Solution

(7) Solve the equation x 4 −14x 2 + 45 = 0 Solution

problem solving involving polynomial equation example

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to [email protected]

We always appreciate your feedback.

Sat Math Practice
SAT Math Worksheets
PEMDAS Rule
BODMAS rule
GEMDAS Order of Operations
Math Calculators
Transformations of Functions
Order of rotational symmetry
Lines of symmetry
Compound Angles
Quantitative Aptitude Tricks
Trigonometric ratio table
Word Problems
Times Table Shortcuts
10th CBSE solution
PSAT Math Preparation
Privacy Policy
Laws of Exponents

Recent Articles

Complex Plane

Apr 14, 24 07:56 AM

Real Analysis

Apr 14, 24 02:50 AM

How to Sketch the Graph of Sinusoidal Function

Apr 13, 24 12:02 PM

HW Guidelines
Study Skills Quiz
Find Local Tutors
Demo MathHelp.com
Join MathHelp.com

Select a Course Below

ACCUPLACER Math
Math Placement Test
PRAXIS Math
+ more tests
5th Grade Math
6th Grade Math
Pre-Algebra
College Pre-Algebra
Introductory Algebra
Intermediate Algebra
College Algebra

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

Content Continues Below

MathHelp.com

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

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.

URL: https://www.purplemath.com/modules/solvpoly3.htm

Page 1 Page 2 Page 3

Standardized Test Prep

College math, homeschool math, share this page.

Terms of Use
About Purplemath
About the Author
Tutoring from PM
Advertising
Linking to PM
Site licencing

Visit Our Profiles

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Polynomial Equations

Learning objectives.

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

Use the Zero Product Property

Solve quadratic equations by factoring
Solve equations with polynomial functions
Solve applications modeled by polynomial equations

Before you get started, take this readiness quiz.

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

${x}^{2}+5x+6=0\phantom{\rule{3em}{0ex}}3{y}^{2}+4y=10\phantom{\rule{3em}{0ex}}64{u}^{2}-81=0\phantom{\rule{3em}{0ex}}n\left(n+1\right)=42$

To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

We will now use the Zero Product Property, to solve a quadratic equation .

$\left(5n-2\right)\left(6n-1\right)=0.$

Set each factor equal to zero.
Solve the linear equations.

Solve Quadratic Equations by Factoring

$a{x}^{2}+bx+c=0.$

Factor the quadratic expression.
Use the Zero Product Property.
Check. Substitute each solution separately into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form .

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

$169{q}^{2}=49.$

We leave the check up to you.

$25{p}^{2}=49.$

$\left(3x-8\right)\left(x-1\right)=3x.$

In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

$3{x}^{2}=12x+63.$

$9{m}^{3}+100m=60{m}^{2}.$

Solve Equations with Polynomial Functions

$f\left(x\right)={x}^{2}+2x-2,$

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function .

$f\left(x\right)=0,$

ⓐ the zeros of the function,

ⓑ any x -intercepts of the graph of the function

ⓐ To find the zeros of the function, we need to find when the function value is 0.

$\begin{array}{cccccc}& & & & & f\left(x\right)=3{x}^{2}+10x-8\hfill \\ \text{Substitute 0 for}\phantom{\rule{0.2em}{0ex}}f\left(x\right).\hfill & & & & & \phantom{\rule{1.2em}{0ex}}0=3{x}^{2}+10x-8\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(x+4\right)\left(3x-2\right)=0\hfill \\ \begin{array}{c}\text{Use the zero product property.}\hfill \\ \text{Solve.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccccc}\hfill x+4& =\hfill & 0\hfill & & & \text{or}\hfill & & & \hfill 3x-2& =\hfill & 0\hfill \\ \hfill x& =\hfill & -4\hfill & & & \text{or}\hfill & & & \hfill x& =\hfill & \frac{2}{3}\hfill \end{array}\hfill \end{array}$

ⓐ the zeros of the function

$x=\frac{5}{2}$

Solve Applications Modeled by Polynomial Equations

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
Solve the equation using appropriate algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a polynomial equation.

The product of two consecutive odd integers is 323. Find the integers.

$\begin{array}{cccccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{We are looking for two consecutive integers.}\hfill \\ \mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{Let}\phantom{\rule{0.2em}{0ex}}n=\text{the first integer.}\hfill \\ & & & & & n+2=\phantom{\rule{0.2em}{0ex}}\text{next consecutive odd integer}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into an equation. Restate the}\hfill \\ \text{problem in a sentence.}\hfill \end{array}\hfill & & & & & \begin{array}{c}\text{The product of the two consecutive odd}\hfill \\ \text{integers is 323.}\hfill \end{array}\hfill \\ & & & & & \phantom{\rule{3.47em}{0ex}}n\left(n+2\right)=323\hfill \\ \mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the equation.}\hfill & & & & & \phantom{\rule{3.57em}{0ex}}{n}^{2}+2n=323\hfill \\ \text{Bring all the terms to one side.}\hfill & & & & & \phantom{\rule{0.85em}{0ex}}{n}^{2}+2n-323=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(n-17\right)\left(n+19\right)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property.}\hfill \\ \text{Solve the equations.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccc}\hfill n-17& =\hfill & 0\hfill & & & & \hfill n+19& =\hfill & 0\hfill \\ \hfill n& =\hfill & 17\hfill & & & & \hfill n& =\hfill & -19\hfill \end{array}\hfill \end{array}$

There are two values for n that are solutions to this problem. So there are two sets of consecutive odd integers that will work.

$\begin{array}{cccccc}\text{If the first integer is}\phantom{\rule{0.2em}{0ex}}n=17\hfill & & & & & \text{If the first integer is}\phantom{\rule{0.2em}{0ex}}n=-19\hfill \\ \text{then the next odd integer is}\hfill & & & & & \text{then the next odd integer is}\hfill \\ \phantom{\rule{11em}{0ex}}n+2\hfill & & & & & \phantom{\rule{11em}{0ex}}n+2\hfill \\ \phantom{\rule{10.5em}{0ex}}17+2\hfill & & & & & \phantom{\rule{10em}{0ex}}\text{−}19+2\hfill \\ \phantom{\rule{11.5em}{0ex}}19\hfill & & & & & \phantom{\rule{11em}{0ex}}\text{−}17\hfill \\ \phantom{\rule{10.5em}{0ex}}17,19\hfill & & & & & \phantom{\rule{10em}{0ex}}\text{−}17,-19\hfill \\ \mathbf{\text{Step 6. Check}}\phantom{\rule{0.2em}{0ex}}\text{the answer.}\hfill & & & & & \\ \text{The results are consecutive odd integers}\hfill & & & & & \\ 17,19\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{−}19,-17.\hfill & & & & & \\ 17·19=323✓\phantom{\rule{4em}{0ex}}\text{−}19\left(\text{−}17\right)=323✓\hfill & & & & & \\ \text{Both pairs of consecutive integers are solutions.}\hfill & & & & & \\ \mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question}\hfill & & & & & \text{The consecutive integers are 17, 19 and}\phantom{\rule{0.2em}{0ex}}\text{−}19,-17.\hfill \end{array}$

The product of two consecutive odd integers is 255. Find the integers.

The product of two consecutive odd integers is 483 Find the integers.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

A rectangular sign has area 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

The width is 5 feet and length is 6 feet.

A rectangular patio has area 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

The length of the patio is 12 feet and the width 15 feet.

$\left({a}^{2}+{b}^{2}={c}^{2}\right).$

We will use this formula to in the next example.

Figure shows a right triangle with the shortest side being x, the second side being x minus 7 and the hypotenuse being 17.

5 feet and 12 feet

24 feet and 25 feet

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

$h\left(t\right)=-16{t}^{2}+64t+80$

ⓐ the zeros of this function which tell us when the ball hits the ground

ⓑ when the ball will be 80 feet above the ground

ⓐ the zeros of this function which tell us when the rock will hit the ocean

ⓑ when the rock will be 160 feet above the ocean.

ⓐ 5 ⓑ 0;3 ⓒ 196

$h\left(t\right)=-16{t}^{2}+32t+128$

ⓐ the zeros of this function which is when the penny will hit the ocean

ⓑ when the penny will be 128 feet above the ocean.

ⓐ 4 ⓑ 0;2 ⓒ 144

Access this online resource for additional instruction and practice with quadratic equations.

Beginning Algebra & Solving Quadratics with the Zero Property

Key Concepts

Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

$a,b,c\phantom{\rule{0.2em}{0ex}}\text{are real numbers and}\phantom{\rule{0.2em}{0ex}}a\ne 0$

Section Exercises

Practice makes perfect.

In the following exercises, solve.

$\left(3a-10\right)\left(2a-7\right)=0$

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

$f\left(x\right)=9{x}^{2}-4$

Solve Applications Modeled by Quadratic Equations

The product of two consecutive odd integers is 143. Find the integers.

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

The product of two consecutive even integers is 288. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

A rectangular carport has area 150 square feet. The height of the carport is five feet less than twice its length. Find the height and the length of the carport.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

$h\left(t\right)=-16{t}^{2}+32t$

ⓐ the zeros of this function which tells us when the rocket will hit the ground. ⓑ the time the rocket will be 16 feet above the ground.

$h\left(t\right)=-16{t}^{2}+32t+48$

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Answers will vary.

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

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

This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve quadratic equations by using the zero product property, solve quadratic equations by factoring and solve applications modeled by quadratic equations.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Greatest common factor and factor by grouping.

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

$12{a}^{2}{b}^{3},15a{b}^{2}$

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

Factor by Grouping

In the following exercises, factor by grouping.

Factor Trinomials

${x}^{2}+bx+c$

In the following exercises, factor completely using trial and error.

${x}^{3}+5{x}^{2}-24x$

In the following exercises, factor.

$2{x}^{2}+9x+4$

Factor using substitution

In the following exercises, factor using substitution.

${x}^{4}-13{x}^{2}-30$

Factor Special Products

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

$25{x}^{2}+30x+9$

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

$81{r}^{2}-25$

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

${a}^{3}-125$

General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

$24{x}^{3}+44{x}^{2}$

In each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

$f\left(x\right)=64{x}^{2}-49$

The product of two consecutive numbers is 399. Find the numbers.

The area of a rectangular shaped patio 432 square feet. The length of the patio is 6 feet more than its width. Find the length and width.

A ladder leans against the wall of a building. The length of the ladder is 9 feet longer than the distance of the bottom of the ladder from the building. The distance of the top of the ladder reaches up the side of the building is 7 feet longer than the distance of the bottom of the ladder from the building. Find the lengths of all three sides of the triangle formed by the ladder leaning against the building.

The lengths are 8, 15, and 17 ft.

Chapter Practice Test

$80{a}^{2}+120{a}^{3}$

In the following exercises, solve

$5{a}^{2}+26a=24$

The product of two consecutive integers is 156. Find the integers.

The area of a rectangular place mat is 168 square inches. Its length is two inches longer than the width. Find the length and width of the placemat.

The width is 12 inches and the length is 14 inches.

Share This Book

school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Download Page (PDF)
Download Full Book (PDF)
Periodic Table
Physics Constants
Scientific Calculator
Reference & Cite
Tools expand_more
Readability

selected template will load here

This action is not available.

1.6: Polynomials and Their Operations

Last updated
Save as PDF
Page ID 6231

Learning Objectives

Identify a polynomial and determine its degree.
Add and subtract polynomials.
Multiply and divide polynomials.

Definitions

A polynomial 112 is a special algebraic expression with terms that consist of real number coefficients and variable factors with whole number exponents. Some examples of polynomials follow:

The degree of a term 113 in a polynomial is defined to be the exponent of the variable, or if there is more than one variable in the term, the degree is the sum of their exponents. Recall that $x^{0} = 1$; any constant term can be written as a product of $x^{0}$ and itself. Hence the degree of a constant term is $0$.

The degree of a polynomial 114 is the largest degree of all of its terms.

Of particular interest are polynomials with one variable 115 , where each term is of the form $a_{n}x^{n}$. Here $a_{n}$ is any real number and $n$ is any whole number. Such polynomials have the standard form:

$a _ { n } x ^ { n } + a _ { n - 1 } x ^ { n - 1 } + \cdots + a _ { 1 } x + a _ { 0 }$

Typically, we arrange terms of polynomials in descending order based on the degree of each term. The leading coefficient 116 is the coefficient of the variable with the highest power, in this case, $a_{n}$.

Example $\PageIndex{1}$:

Write in standard form: $3 x - 4 x ^ { 2 } + 5 x ^ { 3 } + 7 - 2 x ^ { 4 }$.

Since terms are defined to be separated by addition, we write the following:

$\begin{array} { l } { 3 x - 4 x ^ { 2 } + 5 x ^ { 3 } + 7 - 2 x ^ { 4 } } \\ { = 3 x + ( - 4 ) x ^ { 2 } + 5 x ^ { 3 } + 7 + ( - 2 ) x ^ { 4 } } \end{array}$

In this form, we can see that the subtraction in the original corresponds to negative coefficients. Because addition is commutative, we can write the terms in descending order based on the degree as follows:

$\begin{array} { l } { = ( - 2 ) x ^ { 4 } + 5 x ^ { 3 } + ( - 4 ) x ^ { 2 } + 3 x + 7 } \\ { = - 2 x ^ { 4 } + 5 x ^ { 3 } - 4 x ^ { 2 } + 3 x + 7 } \end{array}$

$- 2 x ^ { 4 } + 5 x ^ { 3 } - 4 x ^ { 2 } + 3 x + 7$

We classify polynomials by the number of terms and the degree:

We can further classify polynomials with one variable by their degree:

In this text, we call any polynomial of degree $n ≥ 4$ an $n$th-degree polynomial. In other words, if the degree is $4$, we call the polynomial a fourth-degree polynomial. If the degree is $5$, we call it a fifth-degree polynomial, and so on.

Example $\PageIndex{2}$

State whether the following polynomial is linear or quadratic and give the leading coefficient: $25 + 4 x - x ^ { 2 }$.

The highest power is $2$; therefore, it is a quadratic polynomial. Rewriting in standard form we have

$- x ^ { 2 } + 4 x + 25$

Here $- x ^ { 2 } = - 1 x ^ { 2 }$ and thus the leading coefficient is $−1$.

Quadratic; leading coefficient: $−1$

Adding and Subtracting Polynomials

We begin by simplifying algebraic expressions that look like $+ (a + b)$ or $− (a + b)$. Here, the coefficients are actually implied to be $+1$ and $−1$ respectively and therefore the distributive property applies. Multiply each term within the parentheses by these factors as follows:

$\begin{array} { l } { + ( a + b ) = + 1 ( a + b ) = ( + 1 ) a + ( + 1 ) b = a + b } \\ { - ( a + b ) = - 1 ( a + b ) = ( - 1 ) a + ( - 1 ) b = - a - b } \end{array}$

Use this idea as a means to eliminate parentheses when adding and subtracting polynomials.

Example $\PageIndex{3}$:

Add: $9 x ^ { 2 } + \left( x ^ { 2 } - 5 \right)$.

The property $+ (a + b) = a + b$ allows us to eliminate the parentheses, after which we can then combine like terms.

$\begin{aligned} 9 x ^ { 2 } + \left( x ^ { 2 } - 5 \right) & = 9 x ^ { 2 } + x ^ { 2 } - 5 \\ & = 10 x ^ { 2 } - 5 \end{aligned}$

$10x^{2} − 5$

Example $\PageIndex{4}$:

Add: $\left( 3 x ^ { 2 } y ^ { 2 } - 4 x y + 9 \right) + \left( 2 x ^ { 2 } y ^ { 2 } - 6 x y - 7 \right)$.

Remember that the variable parts have to be exactly the same before we can add the coefficients.

$\begin{array} { l } { \left( 3 x ^ { 2 } y ^ { 2 } - 4 x y + 9 \right) + \left( 2 x ^ { 2 } y ^ { 2 } - 6 x y - 7 \right) } \\ { = \color{Cerulean}{\underline{ 3 x ^ { 2 } y ^ { 2 }}} \color{Black}{-} \color{OliveGreen}{\underline{\underline {4 x y}}} \color{Black}{+ \underline{\underline{\underline{9}}}} + \color{Cerulean}{\underline{2 x ^ { 2 } y ^ { 2 }}} \color{Black}{-} \color{OliveGreen}{\underline{\underline {6 x y}}} \color{Black} {- \underline{\underline{\underline{7}}}}} \\ { = 5 x ^ { 2 } y ^ { 2 } - 10 x y + 2 } \end{array}$

$5 x ^ { 2 } y ^ { 2 } - 10 x y + 2$

When subtracting polynomials, the parentheses become very important.

Example $\PageIndex{5}$:

Subtract: $4 x ^ { 2 } - \left( 3 x ^ { 2 } + 5 x \right)$.

The property $− (a + b) = −a − b$ allows us to remove the parentheses after subtracting each term.

$\begin{aligned} 4 x ^ { 2 } - \left( 3 x ^ { 2 } + 5 x \right) & = 4 x ^ { 2 } - 3 x ^ { 2 } - 5 x \\ & = x ^ { 2 } - 5 x \end{aligned}$

$x^{2} − 5x$

Subtracting a quantity is equivalent to multiplying it by $−1$.

Example $\PageIndex{6}$:

Subtract: $\left( 3 x ^ { 2 } - 2 x y + y ^ { 2 } \right) - \left( 2 x ^ { 2 } - x y + 3 y ^ { 2 } \right)$.

Distribute the $−1$, remove the parentheses, and then combine like terms. Multiplying the terms of a polynomial by $−1$ changes all the signs.

$19157971f267dfed73125fabab9d74ae.png$

$\begin{array} { l } { = 3 x ^ { 2 } - 2 x y + y ^ { 2 } - 2 x ^ { 2 } + x y - 3 y ^ { 2 } } \\ { = x ^ { 2 } - x y - 2 y ^ { 2 } } \end{array}$

$x^{2} − xy − 2y^{2}$

Exercise $\PageIndex{1}$

Subtract: $\left( 7 a ^ { 2 } - 2 a b + b ^ { 2 } \right) - \left( a ^ { 2 } - 2 a b + 5 b ^ { 2 } \right)$.

$6 a ^ { 2 } - 4 b ^ { 2 }$

www.youtube.com/v/IDtREB_PQ3A

Multiplying Polynomials

Use the product rule for exponents, $x ^ { m } \cdot x ^ { n } = x ^ { m + n }$, to multiply a monomial times a polynomial. In other words, when multiplying two expressions with the same base, add the exponents. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For example,

$\begin{aligned} 7 x ^ { 4 } \cdot 8 x ^ { 3 } & = 7 \cdot 8 \cdot x ^ { 4 } \cdot x ^ { 3 } \color{Cerulean} { Commutative \:property } \\ & = 56 x ^ { 4 + 3 } \quad\quad \color {Cerulean} { Product \:rule \:for \:exponents } \\ & = 56 x ^ { 7 } \end{aligned}$

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

Example $\PageIndex{7}$:

Multiply: $5 x y ^ { 2 } \left( 2 x ^ { 2 } y ^ { 2 } - x y + 1 \right)$.

Apply the distributive property and then simplify.

$2603f8e94a90164b1620a009c51f7eaf.png$

$\begin{array} { l } { = \color{Cerulean}{5 x y ^ { 2 }} \color{Black}{\cdot} 2 x ^ { 2 } y ^ { 2 } - \color{Cerulean}{5 x y ^ { 2 }}\color{Black}{ \cdot} x y + \color{Cerulean}{5 x y ^ { 2 }} \color{Black}{ \cdot 1 }} \\ { = 10 x ^ { 3 } y ^ { 4 } - 5 x ^ { 2 } y ^ { 3 } + 5 x y ^ { 2 } } \end{array}$

$10 x ^ { 3 } y ^ { 4 } - 5 x ^ { 2 } y ^ { 3 } + 5 x y ^ { 2 }$

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.

In the same manner that we used the distributive property to distribute a monomial, we use it to distribute a binomial.

$\begin{aligned} \color{Cerulean}{( a + b )}\color{Black}{ ( c + d )} & = \color{Cerulean}{( a + b )}\color{Black}{ \cdot} c + \color{Cerulean}{( a + b )}\color{Black}{ \cdot} d \\ & = a c + b c + a d + b d \\ & = a c + a d + b c + b d \end{aligned}$

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:

$7be734ec5c654af428357e9bd13b2a08.png$

This is often called the FOIL method. Multiply the first, outer, inner, and then last terms.

Example $\PageIndex{8}$:

Multiply: $( 6 x - 1 ) ( 3 x - 5 )$.

Distribute $6x$ and $−1$ and then combine like terms.

$\begin{aligned} ( 6 x - 1 ) ( 3 x - 5 ) & = \color{Cerulean}{6 x}\color{Black}{ \cdot} 3 x - \color{Cerulean}{6 x}\color{Black}{ \cdot} 5 + ( \color{OliveGreen}{- 1}\color{Black}{ )} \cdot 3 x - ( \color{OliveGreen}{- 1}\color{Black}{ )} \cdot 5 \\ & = 18 x ^ { 2 } - 30 x - 3 x + 5 \\ & = 18 x ^ { 2 } - 33 x + 5 \end{aligned}$

$18 x ^ { 2 } - 33 x + 5$

Consider the following two calculations:

This leads us to two formulas that describe perfect square trinomials 124 :

$\begin{array} { l } { ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 } } \\ { ( a - b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 } } \end{array}$

We can use these formulas to quickly square a binomial.

Example $\PageIndex{9}$:

Multiply: $(3x+5)^{2}$

Here $a=3x$ and $b=5$. Apply the formula:

$6f9efdeb4dd8ce181d4304e3f8ada495.png$

$9 x ^ { 2 } + 30 x + 25$

This process should become routine enough to be performed mentally. Our third special product follows:

$\begin{aligned} ( a + b ) ( a - b ) & = a ^ { 2 } - a b + b a - b ^ { 2 } \\ & = a ^ { 2 } \color{red}{- a b + a b}\color{Black}{ -} b ^ { 2 } \\ & = a ^ { 2 } - b ^ { 2 } \end{aligned}$

This product is called difference of squares 125 :

$( a + b ) ( a - b ) = a ^ { 2 } - b ^ { 2 }$

The binomials $(a + b)$ and $(a − b)$ are called conjugate binomials 126 . When multiplying conjugate binomials the middle terms are opposites and their sum is zero; the product is itself a binomial.

Example $\PageIndex{10}$:

Multiply: $(3xy + 1) (3xy − 1)$.

$\begin{aligned} ( 3 x y + 1 ) ( 3 x y - 1 ) & = ( 3 x y ) ^ { 2 } - 3 x y + 3 x y - 1 ^ { 2 } \\ & = 9 x ^ { 2 } y ^ { 2 } - 1 \end{aligned}$

$9x^{2}y^{2} − 1$

Exercise $\PageIndex{2}$

Multiply: $\left( x ^ { 2 } + 5 y ^ { 2 } \right) \left( x ^ { 2 } - 5 y ^ { 2 } \right)$.

$\left( x ^ { 4 } - 25 y ^ { 4 } \right)$

www.youtube.com/v/p7R3FdPp6_s

Example $\PageIndex{11}$:

Multiply: $(5x − 2)^{3}$.

Here we perform one product at a time.

$94bf5b5472fe2ef54e090df103f39d85.png$

$125x^{2} − 150x^{2} + 60x − 8$

Dividing Polynomials

Use the quotient rule for exponents, $\frac { x ^ { m } } { x ^ { n } } = x ^ { m - n }$, to divide a polynomial by a monomial. In other words, when dividing two expressions with the same base, subtract the exponents. In this section, we will assume that all variables in the denominator are nonzero.

Example $\PageIndex{12}$:

Divide: $\frac { 24 x ^ { 7 } y ^ { 5 } } { 8 x ^ { 3 } y ^ { 2 } }$.

Divide the coefficients and apply the quotient rule by subtracting the exponents of the like bases.

$\begin{aligned} \frac { 24 x ^ { 7 } y ^ { 5 } } { 8 x ^ { 3 } y ^ { 2 } } & = \frac { 24 } { 8 } x ^ { 7 - 3 } y ^ { 5 - 2 } \\ & = 3 x ^ { 4 } y ^ { 3 } \end{aligned}$

$3 x ^ { 4 } y ^ { 3 }$

When dividing a polynomial by a monomial, we may treat the monomial as a common denominator and break up the fraction using the following property:

$\frac { a + b } { c } = \frac { a } { c } + \frac { b } { c }$

Applying this property will result in terms that can be treated as quotients of monomials.

Example $\PageIndex{13}$:

Divide: $\frac { - 5 x ^ { 4 } + 25 x ^ { 3 } - 15 x ^ { 2 } } { 5 x ^ { 2 } }$.

Break up the fraction by dividing each term in the numerator by the monomial in the denominator, and then simplify each term.

$\begin{aligned} \frac { - 5 x ^ { 4 } + 25 x ^ { 3 } - 15 x ^ { 2 } } { 5 x ^ { 2 } } & = - \frac { 5 x ^ { 4 } } { 5 x ^ { 2 } } + \frac { 25 x ^ { 3 } } { 5 x ^ { 2 } } - \frac { 15 x ^ { 2 } } { 5 x ^ { 2 } } \\ & = - \frac { 5 } { 5 } x ^ { 4 - 2 } + \frac { 25 } { 5 } x ^ { 3 - 2 } - \frac { 15 } { 5 } x ^ { 2 - 2 } \\ & = - 1 x ^ { 2 } + 5 x ^ { 1 } - 3 x ^ { 0 } \\ & = - x ^ { 2 } + 5 x - 3 \cdot 1 \end{aligned}$

$- x ^ { 2 } + 5 x - 3$

We can check our division by multiplying our answer, the quotient, by the monomial in the denominator, the divisor, to see if we obtain the original numerator, the dividend.

The same technique outlined for dividing by a monomial does not work for polynomials with two or more terms in the denominator. In this section, we will outline a process called polynomial long division 127 , which is based on the division algorithm for real numbers. For the sake of clarity, we will assume that all expressions in the denominator are nonzero.

Example $\PageIndex{14}$:

Divide $\frac { x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4 } { x - 2 }$:

Here $x−2$ is the divisor and $x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4$ is the dividend. To determine the first term of the quotient, divide the leading term of the dividend by the leading term of the divisor.

$597c1846212bb42a41ee60d1ea564951.png$

Multiply the first term of the quotient by the divisor, remembering to distribute, and line up like terms with the dividend.

$5d4e3953ce01091edb8660ea2cbbcd05.png$

Subtract the resulting quantity from the dividend. Take care to subtract both terms.

$59834d557ea8a06825dbda67dbb2956b.png$

Bring down the remaining terms and repeat the process.

$8e8f8648f001e0f2ebe3d06303378e50.png$

Notice that the leading term is eliminated and that the result has a degree that is one less. The complete process is illustrated below:

$12ac7ce3d68c9c0bc218027f83ce92f0.png$

Polynomial long division ends when the degree of the remainder is less than the degree of the divisor. Here, the remainder is $0$. Therefore, the binomial divides the polynomial evenly and the answer is the quotient shown above the division bar.

$\frac { x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4 } { x - 2 } = x ^ { 2 } + 5 x + 2$

To check the answer, multiply the divisor by the quotient to see if you obtain the dividend as illustrated below:

$x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4 = ( x - 2 ) \left( x ^ { 2 } + 5 x + 2 \right)$

This is left to the reader as an exercise.

$x ^ { 2 } + 5 x + 2$

Next, we demonstrate the case where there is a nonzero remainder.

$e7aecc5db5f3a690ad915a7012bd1474.png$

Just as with real numbers, the final answer adds to the quotient the fraction where the remainder is the numerator and the divisor is the denominator. In general, when dividing we have:

$\frac{Dividend}{Divisor}=\color{Cerulean}{Quotient}\color{Black}{+}\frac{\color{OliveGreen}{Remainder}}{\color{Black}{Divisor}}$

If we multiply both sides by the divisor we obtain,

$Dividend=\color{Cerulean}{Quotient}\color{Black}{\times}Divisor +\color{OliveGreen}{Remainder}$

Example $\PageIndex{15}$:

Divide: $\frac { 6 x ^ { 2 } - 5 x + 3 } { 2 x - 1 }$.

Since the denominator is a binomial, begin by setting up polynomial long division.

$907e3b78b0543ab03b8e48c0db9fc92a.png$

To start, determine what monomial times $2x−1$ results in a leading term $6x^{2}$.This is the quotient of the given leading terms: $(6x^{2})÷(2x)=3x$. Multiply $3x$ times the divisor $2x−1$, and line up the result with like terms of the dividend.

$0bf0959dce4e14a83cd7c030a7415b9d.png$

Subtract the result from the dividend and bring down the constant term $+3$.

$6637d797c48b15341303c31af6093ade.png$

Subtracting eliminates the leading term. Multiply $2x−1$ by $−1$ and line up the result.

$e197703db2295db06f984ba307bcf5c2.png$

Subtract again and notice that we are left with a remainder.

$da6de19c485bfb98d24aee7dbea7dae9.png$

The constant term $2$ has degree $0$ and thus the division ends. Therefore,

$\frac { 6 x ^ { 2 } - 5 x + 3 } { 2 x - 1 } = \color{Cerulean}{3 x - 1}\color{Black}{ +} \frac { \color{OliveGreen}{2} } {\color{Black}{ 2 x - 1} }$

To check that this result is correct, we multiply as follows:

$\begin{aligned} \color{Cerulean}{ {quotient }}\color{Black}{ \times} divisor + \color{OliveGreen} {remainder} & \color{Black}{=} \color{Cerulean}{( 3 x - 1 )}\color{Black}{ (} 2 x - 1 ) + \color{OliveGreen}{2} \\ & = 6 x ^ { 2 } - 3 x - 2 x + 1 + 2 \\ & = 6 x ^ { 2 } - 5 x + 2 = dividend\:\: \color{Cerulean}{✓} \end{aligned} $

$3 x - 1 + \frac { 2 } { 2 x - 1 }$

Occasionally, some of the powers of the variables appear to be missing within a polynomial. This can lead to errors when lining up like terms. Therefore, when first learning how to divide polynomials using long division, fill in the missing terms with zero coefficients, called placeholders 128 .

Example $\PageIndex{16}$:

Divide: $\frac { 27 x ^ { 3 } + 64 } { 3 x + 4 }$.

Notice that the binomial in the numerator does not have terms with degree $2$ or $1$. The division is simplified if we rewrite the expression with placeholders:

$27 x ^ { 3 } + 64 = 27 x ^ { 3 } + \color{OliveGreen}{0 x ^ { 2 }}\color{Black}{ +}\color{OliveGreen}{ 0 x}\color{Black}{ +} 64$

Set up polynomial long division:

$3580ae29e84dc2142822da2ab8c2e345.png$

We begin with 27x3÷3x=9x2 and work the rest of the division algorithm.

$c1b11a0e625faf8dfcc54428eb1b21e5.png$

$9 x ^ { 2 } - 12 x + 16$

Example $\PageIndex{17}$:

Divide: $\frac { 3 x ^ { 4 } - 2 x ^ { 3 } + 6 x ^ { 2 } + 23 x - 7 } { x ^ { 2 } - 2 x + 5 }$.

$179294b59c7365cd517ffacf1e4e7086.png$

Begin the process by dividing the leading terms to determine the leading term of the quotient $3x^{4}÷x^{2}=\color{Cerulean}{3x^{2}}$. Take care to distribute and line up the like terms. Continue the process until the remainder has a degree less than $2$.

$6712e898302af732f5e899c4fc6abc3b.png$

The remainder is $x−2$. Write the answer with the remainder:

$\frac { 3 x ^ { 4 } - 2 x ^ { 3 } + 6 x ^ { 2 } + 23 x - 7 } { x ^ { 2 } - 2 x + 5 } = 3 x ^ { 2 } + 4 x - 1 + \frac { x - 2 } { x ^ { 2 } - 2 x + 5 }$

$3 x ^ { 2 } + 4 x - 1 + \frac { x - 2 } { x ^ { 2 } - 2 x + 5 }$

Polynomial long division takes time and practice to master. Work lots of problems and remember that you may check your answers by multiplying the quotient by the divisor (and adding the remainder if present) to obtain the dividend.

Exercise $\PageIndex{3}$

Divide: $\frac { 6 x ^ { 4 } - 13 x ^ { 3 } + 9 x ^ { 2 } - 14 x + 6 } { 3 x - 2 }$.

$2 x ^ { 3 } - 3 x ^ { 2 } + x - 4 - \frac { 2 } { 3 x - 2 }$

www.youtube.com/v/K9NRVMreKAQ

Key Takeaways

Polynomials are special algebraic expressions where the terms are the products of real numbers and variables with whole number exponents.
The degree of a polynomial with one variable is the largest exponent of the variable found in any term. In addition, the terms of a polynomial are typically arranged in descending order based on the degree of each term.
When adding polynomials, remove the associated parentheses and then combine like terms. When subtracting polynomials, distribute the $−1$, remove the parentheses, and then combine like terms.
To multiply polynomials apply the distributive property; multiply each term in the first polynomial with each term in the second polynomial. Then combine like terms.
When dividing by a monomial, divide all terms in the numerator by the monomial and then simplify each term.
When dividing a polynomial by another polynomial, apply the division algorithm.

Exercise $\PageIndex{4}$

Write the given polynomials in standard form.

$1 − x − x^{2}$
$y − 5 + y^{2}$
$y − 3y^{2} + 5 − y^{3}$
$8 − 12a^{2} + a^{3} − a$
$2 − x^{2} + 6x − 5x^{3} + x^{4}$
$a^{3} − 5 + a^{2} + 2a^{4} − a^{5} + 6a$

1. $- x ^ { 2 } - x + 1$

3. $- y ^ { 3 } - 3 y ^ { 2 } + y + 5$

5. $x ^ { 4 } - 5 x ^ { 3 } - x ^ { 2 } + 6 x + 2$

Exercise $\PageIndex{5}$

Classify the given polynomial as a monomial, binomial, or trinomial and state the degree.

$x^{2} − x + 2$
$5 − 10x^{3}$
$x^{2}y^{2} + 5xy − 6$
$−2x^{3}y^{2}$
$x^{4} − 1$

1. 7. Trinomial; degree $2$

3. Trinomial; degree $4$

5. Binomial; degree $4$

Exercise $\PageIndex{6}$

State whether the polynomial is linear or quadratic and give the leading coefficient.

$1 − 9x^{2}$
$10x^{2}$
$2x − 3$
$5x^{2} + 3x − 1$
$x − 1$
$x − 6 − 2x^{2}$
$1 − 5x$

1. Quadratic, $−9$

3. Linear, $2$

5. Quadratic, $5$

7. Quadratic, $−2$

Exercise $\PageIndex{7}$

$\left( 5 x ^ { 2 } - 3 x - 2 \right) + \left( 2 x ^ { 2 } - 6 x + 7 \right)$
$\left( x ^ { 2 } + 7 x - 12 \right) + \left( 2 x ^ { 2 } - x + 3 \right)$
$\left( x ^ { 2 } + 5 x + 10 \right) + \left( x ^ { 2 } - 10 \right)$
$\left( x ^ { 2 } - 1 \right) + ( 4 x + 2 )$
$\left( 10 x ^ { 2 } + 3 x - 2 \right) - \left( x ^ { 2 } - 6 x + 1 \right)$
$\left( x ^ { 2 } - 3 x - 8 \right) - \left( 2 x ^ { 2 } - 3 x - 8 \right)$
$\left( \frac { 2 } { 3 } x ^ { 2 } + \frac { 3 } { 4 } x - 1 \right) - \left( \frac { 1 } { 6 } x ^ { 2 } + \frac { 5 } { 2 } x - \frac { 1 } { 2 } \right)$
$\left( \frac { 4 } { 5 } x ^ { 2 } - \frac { 5 } { 8 } x + \frac { 10 } { 6 } \right) - \left( \frac { 3 } { 10 } x ^ { 2 } - \frac { 2 } { 3 } x + \frac { 3 } { 5 } \right)$
$\left( x ^ { 2 } y ^ { 2 } + 7 x y - 5 \right) - \left( 2 x ^ { 2 } y ^ { 2 } + 5 x y - 4 \right)$
$\left( x ^ { 2 } - y ^ { 2 } \right) - \left( x ^ { 2 } + 6 x y + y ^ { 2 } \right)$
$\left( a ^ { 2 } b ^ { 2 } + 5 a b - 2 \right) + ( 7 a b - 2 ) - \left( 4 - a ^ { 2 } b ^ { 2 } \right)$
$\left( a ^ { 2 } + 9 a b - 6 b ^ { 2 } \right) - \left( a ^ { 2 } - b ^ { 2 } \right) + 7 a b$
$\left( 10 x ^ { 2 } y - 8 x y + 5 x y ^ { 2 } \right) - \left( x ^ { 2 } y - 4 x y \right) + \left( x y ^ { 2 } + 4 x y \right)$
$\left( 2 m ^ { 2 } n - 6 m n + 9 m n ^ { 2 } \right) - \left( m ^ { 2 } n + 10 m n \right) - m ^ { 2 } n$
$\left( 8 x ^ { 2 } y ^ { 2 } - 5 x y + 2 \right) - \left( x ^ { 2 } y ^ { 2 } + 5 \right) + ( 2 x y - 3 )$
$\left( x ^ { 2 } - y ^ { 2 } \right) - \left( 5 x ^ { 2 } - 2 x y - y ^ { 2 } \right) - \left( x ^ { 2 } - 7 x y \right)$
$\left( \frac { 1 } { 6 } a ^ { 2 } - 2 a b + \frac { 3 } { 4 } b ^ { 2 } \right) - \left( \frac { 5 } { 3 } a ^ { 2 } + \frac { 4 } { 5 } b ^ { 2 } \right) + \frac { 11 } { 8 } a b$
$\left( \frac { 5 } { 2 } x ^ { 2 } - 2 y ^ { 2 } \right) - \left( \frac { 7 } { 5 } x ^ { 2 } - \frac { 1 } { 2 } x y + \frac { 7 } { 3 } y ^ { 2 } \right) - \frac { 1 } { 2 } x y$
$\left( x ^ { 2 n } + 5 x ^ { n } - 2 \right) + \left( 2 x ^ { 2 n } - 3 x ^ { n } - 1 \right)$
$\left( 7 x ^ { 2 n } - x ^ { n } + 5 \right) - \left( 6 x ^ { 2 n } - x ^ { n } - 8 \right)$
Subtract $4y − 3$ from $y^{2} + 7y − 10$.
Subtract $x^{2} + 3x − 2$ from $2x^{2} + 4x − 1$.
A right circular cylinder has a height that is equal to the radius of the base, $h = r$. Find a formula for the surface area in terms of $h$.
A rectangular solid has a width that is twice the height and a length that is $3$ times that of the height. Find a formula for the surface area in terms of the height.

1. $7 x ^ { 2 } - 9 x + 5$

3. $2 x ^ { 2 } + 5 x$

5. $9 x ^ { 2 } + 9 x - 3$

7. $\frac { 1 } { 2 } x ^ { 2 } - \frac { 7 } { 4 } x - \frac { 1 } { 2 }$

9. $- x ^ { 2 } y ^ { 2 } + 2 x y - 1$

11. $2 a ^ { 2 } b ^ { 2 } + 12 a b - 8$

13. $9 x ^ { 2 } y + 6 x y ^ { 2 }$

15. $7 x ^ { 2 } y ^ { 2 } - 3 x y - 6$

17. $- \frac { 3 } { 2 } a ^ { 2 } - \frac { 5 } { 8 } a b - \frac { 1 } { 20 } b ^ { 2 }$

19. $- \frac { 3 } { 2 } a ^ { 2 } - \frac { 5 } { 8 } a b - \frac { 1 } { 20 } b ^ { 2 }$

21. $y ^ { 2 } + 3 y - 7$

23. $S A = 4 \pi h ^ { 2 }$

Exercise $\PageIndex{8}$

$- 8 x ^ { 2 } \cdot 2 x$
$- 10 x ^ { 2 } y \cdot 5 x ^ { 3 } y ^ { 2 }$
$2 x ( 5 x - 1 )$
$- 4 x ( 3 x - 5 )$
$7 x ^ { 2 } ( 2 x - 6 )$
$- 3 x ^ { 2 } \left( x ^ { 2 } - x + 3 \right)$
$- 5 y ^ { 4 } \left( y ^ { 2 } - 2 y + 3 \right)$
$\frac { 5 } { 2 } a ^ { 3 } \left( 24 a ^ { 2 } - 6 a + 4 \right)$
$2 x y \left( x ^ { 2 } - 7 x y + y ^ { 2 } \right)$
$- 2 a ^ { 2 } b \left( a ^ { 2 } - 3 a b + 5 b ^ { 2 } \right)$
$x ^ { n } \left( x ^ { 2 } + x + 1 \right)$
$x ^ { n } \left( x ^ { 2 n } - x ^ { n } - 1 \right)$
$( x + 4 ) ( x - 5 )$
$( x - 7 ) ( x - 6 )$
$( 2 x - 3 ) ( 3 x - 1 )$
$( 9 x + 1 ) ( 3 x + 2 )$
$\left( 3 x ^ { 2 } - y ^ { 2 } \right) \left( x ^ { 2 } - 5 y ^ { 2 } \right)$
$\left( 5 y ^ { 2 } - x ^ { 2 } \right) \left( 2 y ^ { 2 } - 3 x ^ { 2 } \right)$
$( 3 x + 5 ) ( 3 x - 5 )$
$( x + 6 ) ( x - 6 )$
$\left( a ^ { 2 } - b ^ { 2 } \right) \left( a ^ { 2 } + b ^ { 2 } \right)$
$( a b + 7 ) ( a b - 7 )$
$\left( 4 x - 5 y ^ { 2 } \right) \left( 3 x ^ { 2 } - y \right)$
$( x y + 5 ) ( x - y )$
$( x - 5 ) \left( x ^ { 2 } - 3 x + 8 \right)$
$( 2 x - 7 ) \left( 3 x ^ { 2 } - x + 1 \right)$
$\left( x ^ { 2 } + 7 x - 1 \right) \left( 2 x ^ { 2 } - 3 x - 1 \right)$
$\left( 4 x ^ { 2 } - x + 6 \right) \left( 5 x ^ { 2 } - 4 x - 3 \right)$
$( x + 8 ) ^ { 2 }$
$( x - 3 ) ^ { 2 }$
$( 2 x - 5 ) ^ { 2 }$
$( 3 x + 1 ) ^ { 2 }$
$( a - 3 b ) ^ { 2 }$
$( 7 a - b ) ^ { 2 }$
$\left( x ^ { 2 } + 2 y ^ { 2 } \right) ^ { 2 }$
$\left( x ^ { 2 } - 6 y \right) ^ { 2 }$
$\left( a ^ { 2 } - a + 5 \right) ^ { 2 }$
$\left( x ^ { 2 } - 3 x - 1 \right) ^ { 2 }$
$( x - 3 ) ^ { 3 }$
$( x + 2 ) ^ { 3 }$
$( 3x + 1 ) ^ { 3 }$
$( 2x - 3 ) ^ { 3 }$
$( x + 2 ) ^ { 4 }$
$( x - 3 ) ^ { 4 }$
$( 2x - 1 ) ^ { 4 }$
$( 3x - 1 ) ^ { 4 }$
$\left( x ^ { 2 n } + 5 \right) \left( x ^ { 2 n } - 5 \right)$
$\left( x ^ { n } - 1 \right) \left( x ^ { 2 n } + 4 x ^ { n } - 3 \right)$
$\left( x ^ { 2 n } - 1 \right) ^ { 2 }$
$\left( x ^ { 3 n } + 1 \right) ^ { 2 }$
Find the product of $3x-2$ and $x^{2}-5x-2$.
Find the product of $x^{2}+4$ and $x^{3}-1$.
Each side of a square measures $3x^{3}$ units. Determine the area in terms of $x$.
Each edge of a cube measures $2x^{2}$ units. Determine the volume in terms of $x$.

1. $-16x^{3}$

3. $10 x ^ { 2 } - 2 x$

5. $14 x ^ { 3 } - 42 x ^ { 2 }$

7. $- 5 y ^ { 6 } + 10 y ^ { 5 } - 15 y ^ { 4 }$

9. $2 x ^ { 3 } y - 14 x ^ { 2 } y ^ { 2 } + 2 x y ^ { 3 }$

11. $x ^ { n + 2 } + x ^ { n + 1 } + x ^ { n }$

13. $x ^ { 2 } - x - 20$

15. $6 x ^ { 2 } - 11 x + 3$

17. $3 x ^ { 4 } - 16 x ^ { 2 } y ^ { 2 } + 5 y ^ { 4 }$

19. $9 x ^ { 2 } - 25$

21. $a ^ { 4 } - b ^ { 4 }$

23. $12 x ^ { 3 } - 15 x ^ { 2 } y ^ { 2 } - 4 x y + 5 y ^ { 3 }$

25. $x ^ { 3 } - 8 x ^ { 2 } + 23 x - 40$

27. $2 x ^ { 4 } + 11 x ^ { 3 } - 24 x ^ { 2 } - 4 x + 1$

29. $x ^ { 2 } + 16 x + 64$

31. $4 x ^ { 2 } - 20 x + 25$

33. $a ^ { 2 } - 6 a b + 9 b ^ { 2 }$

35. $x ^ { 4 } + 4 x ^ { 2 } y ^ { 2 } + 4 y ^ { 4 }$

37. $a ^ { 4 } - 2 a ^ { 3 } + 11 a - 10 a + 25$

39. $x ^ { 3 } - 9 x ^ { 2 } + 27 x - 27$

41. $27 x ^ { 3 } + 27 x ^ { 2 } + 9 x + 1$

43. $x ^ { 4 } + 8 x ^ { 3 } + 24 x ^ { 2 } + 32 x + 16$

45. $16 x ^ { 4 } - 32 x ^ { 3 } + 24 x ^ { 2 } - 8 x + 1$

47. $x ^ { 4 n } - 25$

49. $x ^ { 4 n } - 2 x ^ { 2 n } + 1$

51. $3 x ^ { 3 } - 17 x ^ { 2 } + 4 x + 4$

53. $9 x ^ { 6 }$ square units

Exercise $\PageIndex{9}$

$\frac { 125 x ^ { 5 } y ^ { 2 } } { 25 x ^ { 4 } y ^ { 2 } }$
$\frac { 256 x ^ { 2 } y ^ { 3 } z ^ { 5 } } { 64 x ^ { 2 } y z ^ { 2 } }$
$\frac { 20 x ^ { 3 } - 12 x ^ { 2 } + 4 x } { 4 x }$
$\frac { 15 x ^ { 4 } - 75 x ^ { 3 } + 18 x ^ { 2 } } { 3 x ^ { 2 } }$
$\frac { 12 a ^ { 2 } b + 28 a b ^ { 2 } - 4 a b } { 4 a b }$
$\frac { - 2 a ^ { 4 } b ^ { 3 } + 16 a ^ { 2 } b ^ { 2 } + 8 a b ^ { 3 } } { 2 a b ^ { 2 } }$
$\frac { x ^ { 3 } + x ^ { 2 } - 3 x + 9 } { x + 3 }$
$\frac { x ^ { 3 } - 4 x ^ { 2 } - 9 x + 20 } { x - 5 }$
$\frac { 6 x ^ { 3 } - 11 x ^ { 2 } + 7 x - 6 } { 2 x - 3 }$
$\frac { 9 x ^ { 3 } - 9 x ^ { 2 } - x + 1 } { 3 x - 1 }$
$\frac { 16 x ^ { 3 } + 8 x ^ { 2 } - 39 x + 17 } { 4 x - 3 }$
$\frac { 12 x ^ { 3 } - 56 x ^ { 2 } + 55 x + 30 } { 2 x - 5 }$
$\frac { 6 x ^ { 4 } + 13 x ^ { 3 } - 9 x ^ { 2 } - x + 6 } { 3 x + 2 }$
$\frac { 25 x ^ { 4 } - 10 x ^ { 3 } + 11 x ^ { 2 } - 7 x + 1 } { 5 x - 1 }$
$\frac { 20 x ^ { 4 } + 12 x ^ { 3 } + 9 x ^ { 2 } + 10 x + 5 } { 2 x + 1 }$
$\frac { 25 x ^ { 4 } - 45 x ^ { 3 } - 26 x ^ { 2 } + 36 x - 11 } { 5 x - 2 }$
$\frac { 3 x ^ { 4 } + x ^ { 2 } - 1 } { x - 2 }$
$\frac { x ^ { 4 } + x - 3 } { x + 3 }$
$\frac { x ^ { 3 } - 10 } { x - 2 }$
$\frac { x ^ { 3 } + 15 } { x + 3 }$
$\frac { y ^ { 5 } + 1 } { y + 1 }$
$\frac { y ^ { 6 } + 1 } { y + 1 }$
$\frac { x ^ { 4 } - 4 x ^ { 3 } + 6 x ^ { 2 } - 7 x - 1 } { x ^ { 2 } - x + 2 }$
$\frac { 6 x ^ { 4 } + x ^ { 3 } - 2 x ^ { 2 } + 2 x + 4 } { 3 x ^ { 2 } - x + 1 }$
$\frac { 2 x ^ { 3 } - 7 x ^ { 2 } + 8 x - 3 } { x ^ { 2 } - 2 x + 1 }$
$\frac { 2 x ^ { 4 } + 3 x ^ { 3 } - 6 x ^ { 2 } - 4 x + 3 } { x ^ { 2 } + x - 3 }$
$\frac { x ^ { 4 } + 4 x ^ { 3 } - 2 x ^ { 2 } - 4 x + 1 } { x ^ { 2 } - 1 }$
$\frac { x ^ { 4 } + x - 1 } { x ^ { 2 } + 1 }$
$\frac { x ^ { 3 } + 6 x ^ { 2 } y + 4 x y ^ { 2 } - y ^ { 3 } } { x + y }$
$\frac { 2 x ^ { 3 } - 3 x ^ { 2 } y + 4 x y ^ { 2 } - 3 y ^ { 3 } } { x - y }$
$\frac { 8 a ^ { 3 } - b ^ { 3 } } { 2 a - b }$
$\frac { a ^ { 3 } + 27 b ^ { 3 } } { a + 3 b }$
Find the quotient of $10 x ^ { 2 } - 11 x + 3$ and $2x-1$.
Find the quotient of $12 x ^ { 2 } + x - 11$ and $3x-2$.

3. $5 x ^ { 2 } - 3 x + 1$

5. $3 a + 7 b - 1$

7. $x ^ { 2 } - 2 x + 3$

9. $3 x ^ { 2 } - x + 2$

11. $4 x ^ { 2 } + 5 x - 6 - \frac { 1 } { 4 x - 3 }$

13. $2 x ^ { 3 } + 3 x ^ { 2 } - 5 x + 3$

15. $10 x ^ { 3 } + x ^ { 2 } + 4 x + 3 + \frac { 2 } { 2 x + 1 }$

17. $3 x ^ { 3 } + 6 x ^ { 2 } + 13 x + 26 + \frac { 51 } { x - 2 }$

19. $x ^ { 2 } + 2 x + 4 - \frac { 2 } { x - 2 }$

21. $y ^ { 4 } - y ^ { 3 } + y ^ { 2 } - y + 1$

23. $x ^ { 2 } - 3 x + 1 - \frac { 3 } { x ^ { 2 } - x + 2 }$

25. $2 x - 3$

27. $x ^ { 2 } + 4 x - 1$

29. $x ^ { 2 } + 5 x y - y ^ { 2 }$

31. $4 a ^ { 2 } + 2 a b + b ^ { 2 }$

33. $5x-3$

112 An algebraic expression consisting of terms with real number coefficients and variables with whole number exponents.

113 The exponent of the variable. If there is more than one variable in the term, the degree of the term is the sum their exponents.

114 The largest degree of all of its terms.

115 A polynomial where each term has the form $a_{n}x^{n}$, where $a_{n}$ is any real number and $n$ is any whole number.

116 The coefficient of the term with the largest degree.

117 Polynomial with one term.

118 Polynomial with two terms.

119 Polynomial with three terms.

120 A polynomial with degree $0$.

121 A polynomial with degree $1$.

122 A polynomial with degree $2$.

123 A polynomial with degree $3$.

124 The trinomials obtained by squaring the binomials $(a + b)^{2} = a^{2} + 2ab + b^{2}$ and $(a − b)^{2} = a^{2} − 2ab + b^{2}$.

125 The special product obtained by multiplying conjugate binomials $( a + b ) ( a - b ) = a ^ { 2 } - b ^ { 2 }$.

126 The binomials $(a + b)$ and $(a − b)$.

127 The process of dividing two polynomials using the division algorithm.

128 Terms with zero coefficients used to fill in all missing exponents within a polynomial.

Grade 10 Mathematics Module: Solving Problems Involving Polynomial Functions

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

The SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pretest is provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the posttest to self-check your learning. Answer Key is provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

This module was designed and written with you in mind. It is here to help you solve problems involving polynomial functions applying the concepts learned in the previous modules. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course but the order in which you read and answer this module is dependent on your ability.

After going through this module, you are expected to solve problems involving polynomial functions.