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Factoring Polynomials

- Chapter 6

Chapter Sections

6.1 Greatest Common Factor and Factoring by

Grouping 6.2 Factoring Trinomials of the Form

x2 bx c 6.3 Factoring Trinomials of the

Form ax2 bx c, a ? 1 6.4 Factoring Special

Products 6.5 Summary of Factoring

Techniques 6.6 Solving Polynomial Equations by

Factoring 6.7 Modeling and Solving Problems

with Quadratic Equations

Greatest Common Factor and Factoring by Grouping

- Section 6.1

Factors

4 9 36 3(x 2) 3x 6 (2x 7)(3x 5)

6x2 11x 35

The expressions on the left side are called

factors of the expression on the right side.

The greatest common factor (GCF) of a list of

algebraic expressions is the largest expression

that divides evenly into all the expressions.

Greatest Common Factor

Finding the Greatest Common Factor Step 1 Write

each number as a product of prime factors. Step

2 Determine the common prime factors. Step 3

Find the product of the common factors found in

Step 2. This number is the GCF.

Example Find the GCF of 16 and 20.

16 2 2 2 2

20 2 2 5

The GCF is 2 2 4.

Greatest Common Factor

Example Find the GCF of 60, 75, and 135.

60 2 2 3 5

75 3 5 5

135 3 3 3 5

The GCF is 3 5 15.

The GCF as a Binomial

Example Find the GCF of 6(x y) and 15(x y)3.

6(x y) 2 3 (x y)

15(x y)3 3 5 (x y) (x y) (x

y)

The GCF is 3 (x y) 3(x y).

Finding the GCF

Steps to Find the Greatest Common Factor of Two

or More Expressions Step 1 Find the GCF of the

coefficients of each variable expression. Step 2

For each variable expression, determine the

smallest exponent that the variable expression is

raised to. Step 3 Find the product of the

common factors found in Steps 1 and 2. This

expression is the GCF.

Factoring Polynomials

Steps to Factor a Polynomial Using the GCF Step

1 Identify the GCF of the terms that make up

the polynomial. Step 2 Rewrite each term as the

product of the GCF and the remaining factor. Step

3 Use the Distributive Property in reverse to

factor out the GCF. Step 4 Check Use the

Distributive Property to verify that the

factorization is correct.

Factoring Polynomials

Example Factor the trinomial 36a6 45a4 18a2

by factoring out the GCF.

Step 1 Find the GCF.

GCF 9a2

Step 2 Rewrite each term as the product of the

GCF and the remaining term.

36a6 45a4 18a2 9a2 4a4 9a2 5a2 9a2

2

Step 3 Factor out the GCF.

36a6 45a4 18a2 9a2(4a4 5a2 2)

Step 4 Check.

9a2(4a4 5a2 2) 36a6 45a4 18a2

?

Factoring Out a Negative Number

Example Factor 3x6 9x4 18x by factoring

out the GCF

Step 1 Find the GCF.

GCF 3x

Step 2 Rewrite each term as the product of the

GCF and the remaining term.

3x6 9x4 18x 3x x5 ( 3x)( 3x3)

( 3x) 6

Step 3 Factor out the GCF.

3x6 9x4 18x 3x(x5 3x3 6)

Step 4 Check.

3x(x5 3x3 6) 3x6 9x4 18x

?

Factoring Out a Binomial

Example Factor out the greatest common binomial

factor 6(3x y) z(3x y)

6(3x y) z(3x y) (3x y)(6 z)

Check

(3x y)(6 z) 6(3x y) z(3x y)

?

Factoring by Grouping

Steps to Factor by Grouping Step 1 Group the

terms with common factors. Step 2 In each

grouping, factor out the greatest common

factor. Step 3 Factor out the common factor

that remains. Step 4 Check your work.

Factoring by Grouping

Example Factor by grouping x2 7x 3x 21

x2 7x 3x 21

x2 7x 3x 21 x(x 7)

3(x 7)

(x 3)(x 7)

Factor out the common factor x 7.

(x 3)(x 7) x2 7x 3x 21

Check

?

Factoring by Grouping

Example Factor by grouping xy 4x 3y 12

Step 1 Group the terms with the common factors.

(xy 4x) ( 3y 12)

Step 2 Factor out the common factor in each

group.

xy 4x 3y 12 x(y 4) ( 3)(y 4)

Step 3 Factor out the common factor that

remains.

(x 3)(y 4)

xy 4x 3y 12 (x 3)(y 4)

Step 4 Check.

(x 3)(y 4) xy 4x 3y 12

?

Factoring Trinomials of the Form x2 bx c

- Section 6.2

Quadratic Trinomials

A quadratic trinomial is a polynomial of the form

ax2 bx c, a ? 0 where a represents

the coefficient of the squared (second degree)

term, b represents the coefficient of the linear

(first degree) term and c represents the

constant.

3x2 4x 7

8a2 24a 10

6c2 c 25

When the trinomial is written in standard form

(or descending order of degree), the coefficient

of the squared term is called the leading

coefficient.

Trinomials of the Form x2 bx c

Factoring a Trinomial of the Form x2 bx

c Step 1 Find the pair of integers whose

product is c and whose sum is b. That is,

determine m and n such that mn c and m n

b. Step 2 Write x2 bx c (x m)(x

n). Step 3 Check your work by multiplying the

binomials using the FOIL method.

The coefficient of x is the sum of the two

numbers.

x2 11x 24 (x 3)(x 8)

The last term is the

product of the two numbers.

Trinomials of the Form x2 bx c

Example Factor x2 8x 15

x2 8x 15 (x ?)(x ?)

3

5

___ ___ 15

Find two numbers that we can multiply together to

get 15 and add together to get 8.

3

5

___ ___ 8

x2 8x 15 (x 3)(x 5)

Check

?

(x 3)(x 5) x2 8x 15

Trinomials of the Form x2 bx c, c lt 0

Example Factor x2 x 42

x2 x 42

(x ?)(x ?)

One factor will be positive and one will be

negative.

(? 6)

7

____ ____ ? 42

Find two numbers that we can multiply together to

get 42 and add together to get 1.

(? 6)

7

____ ____ 1

x2 x 42 (x 7)(x ? 6)

Check

?

(x 7)(x 6 ) x2 x 42

Trinomials of the Form x2 bx c

The following table summarizes the four forms for

factoring a quadratic trinomial in the form x2

bx c.

Prime Polynomials

A polynomial that cannot be written as the

product of two other polynomials (other than 1 or

1) is said to be a prime polynomial.

Example Factor 5x2 ? x ? 2

The Factors of 5 are

The Factors of 2 are

1 and 5

1 and 2

Possible Factors

Middle Term

(x 1)(5x 2)

3x

9x

(x 2)(5x 1)

The polynomial 5x2 ? x ? 2 is prime.

Trinomials with a Common Factor

Example Factor 2x2 32x 96

2x2 32x 96

2(x2 16x 48)

The common factor of 2 can be factored out.

x2 16x 48

(x ?)(x ?)

(? 4)

(? 12)

____ ____ 48

Find two numbers that we can multiply together to

get 48 and add together to get 16.

(? 4)

(? 12)

____ ____ 16

2x2 32x 96 2(x 12)(x ? 4)

Check

2(x 12)(x ? 4) 2(x2 16x 48)

?

2x2 32x 96

Negative Leading Coefficients

Example Factor x2 12x 36

It is easier to factor out a GCF of 1.

x2 12x 36 1(x2 12x 36)

1(x2 12x 36) 1(x ?)(x ?)

6

6

___ ___ 36

Find two numbers that we can multiply together to

get 36 and add together to get 12.

6

6

___ ___ 12

1(x2 12x 36) 1(x 6)(x 6)

(x 6)2

Check

(x 6)2 1(x 6)(x 6)

?

x2 12x 36

Factoring Trinomials of the Form

ax2 bx c, a ? 1

- Section 6.3

Trinomials of the Form ax2 bx c, a ? 1

When factoring trinomials of the form ax2 bx

c where a is not equal to 1, there are two

methods that can be used 1. Trial and error

2. Factoring by grouping

Factoring ax2 bx c, a ? 1 Using Trial and

Error a, b, and c Have No Common Factors Step 1

List the possibilities for the first terms of

each binomial whose product is ax2. Step 2 List

the possibilities for the last terms of each

binomial whose product is c. Step 3 Write out

all the combinations of factors found in Steps 1

and 2. Multiply the binomials out until a

product is found that equals the trinomial.

( __x )( __x ) ax2 bx c

The Trial and Error Method

Example Factor by trial and error 3x2 4x 1

The Factors of 3 are

The Factors of 1 are

1 and 3

1 and 1

Possible Factors

Middle Term

(x 1)(3x 1)

4x

(x 1)(3x 1) 3x2 4x 1

Check

?

The Trial and Error Method

Example Factor 5x2 2x ? 7

The sign of one factor will be positive, the sign

of the other factor will be negative.

The Factors of 5 Are

The Factors of 7 Are

1 and 5

1 and 7

Possible Factors

Middle Term

(x 1)(5x ? 7)

2x

2x

(x ? 1)(5x 7)

5x2 2x ? 7 (x 1)(5x 7)

(x 1)(5x 7) 5x2 2x ? 7

Check

?

The Trial and Error Method

Example Factor by trial and error 9x2 ? 13x

4

The Factors of 9 are

The Factors of 4 are

1 and 9

1 and 4

2 and 2

3 and 3

Possible Factors

Middle Term

(x 1)(9x 4)

13x

20x

(x 2)(9x 2)

15x

(3x 1)(3x 4)

12x

(3x 2)(3x 2)

9x2 ? 13x 4 (x 1)(9x 4)

Check

?

Factoring by Grouping

Factoring ax2 bx c, a ? 1 by Grouping a, b,

and c Have No Common Factors Step 1 Find the

value of ac. Step 2 Find the pair of integers

whose product is ac and whose sum is b. (Find m

and n so that mn ac and m n b.) Step 3

Write ax2 bx c ax2 mx nx c. Step 4

Factor the expression in Step 3 by grouping. Step

5 Check Multiply out the factored form.

Factoring by Grouping

Example Factor by grouping 2x2 9x 4

The value of ac 8.

Find two numbers that we can multiply together to

get 8 and add together to get 9.

1

8

___ ___ 8

1

8

___ ___ 9

2x2 9x 4 2x2 x 8x 4

Rewrite 9x as x 8x.

x(2x 1) 4(2x 1)

Factor by grouping.

(2x 1)(x 4)

Factor out (2x 1).

(2x 1)(x 4)

2x2 8x 1x 4

Check

2x2 9x 4

?

Factoring by Grouping

Example Factor by grouping 8x2 10x ? 3

The value of ac is ? 24.

(? 2)

12

____ ____ ? 24

Find two numbers that we can multiply together to

get ? 24 and add together to get 10.

(? 2)

12

____ ____ 10

8x2 10x ? 3 8x2 12x ? 2x ? 3

Rewrite 10x as 12x (?2x).

4x(2x 3) ? 1(2x 3)

Factor by grouping.

(2x 3)(4x ? 1)

Factor out (2x 3).

(2x 3)(4x ? 1) 8x2 10x ? 3

?

Check

Factoring Special Products

- Section 6.4

Perfect Square Trinomials

Perfect Square Trinomials a2 2ab b2 (a

b)2 a2 2ab b2 (a ? b)2

In order for a polynomial to be a perfect

square trinomial, two conditions must be

satisfied 1. The first and last terms must

be perfect squares. 2. The middle term must

equal 2 or 2 times the product of the

expressions being squared in the first and last

term.

Perfect Square Trinomials

Example Factor the following

The first term, x2, and third term, 16, are

perfect squares.

a.) x2 8x 16

The middle term, 8x, is 2 times the product of x

and 4.

x2 8x 16 (x 4)2

b.) 9x4 ? 30x2z 25z2

9x4 ? 30x2z 25z2 (3x2 5z)2

The Difference of Two Squares

Difference of Two Squares a2 b2 (a b)(a ? b)

Example Factor the following

a.) 3x2 ? 27

3x2 ? 27 3(x2 9)

3 is a common factor.

3(x 3)(x 3)

b.) 4x2 ? 25y4

4x2 ? 25y4 (2x 5y2)(2x 5y2)

The Sum of Two Cubes

The Sum of Two Cubes a3 b3 (a b) (a2 ? ab

b2)

Example Factor 27x3 125

a 3x

b 5

27x3 125 (3x)3 53

(3x 5) (3x)2 ? 3x5 52

(3x 5)(9x2 ? 15x 25)

The Difference of Two Cubes

The Difference of Two Cubes a3 ? b3 (a ? b) (a2

ab b2)

Example Factor x3 ? 64

a x

b 4

x3 ? 64 x3 43

(x ? 4) (x2 4x 42)

(x ? 4) (x2 4x 16)

Summary of Factoring Techniques

- Section 6.5

Steps for Factoring

Steps for Factoring Step 1 Is there a greatest

common factor? If so, factor out the GCF. Step

2 Count the number of terms. Step 3 (a) 2

terms (Binomials)

- Is it the difference of two squares? If so,
- a2 b2 (a b)(a b)

- Is it the difference of two cubes? If so,
- a3 b3 (a b)(a2 ab b2)

- Is it the sum of two cubes? If so,
- a3 b3 (a b)(a2 ab b2)

Continued.

Steps for Factoring

Steps for Factoring (Continued) (b) 3 terms

(Trinomials)

- Is it a perfect square trinomial? If so,
- a2 2ab b2 (a b)2 or a2 2ab b2 (a

b)2

- Is the coefficient of the square term 1? If so,
- x2 bx c (x m)(x n) where mn c and m

n b

- Is the coefficient of the square term ? 1? If

so, - a. Use factoring by grouping
- b. Use trial and error

(c) 4 terms

- Use factoring by grouping

Step 4 Check each factor to determine if it can

be factored again.

Step 5 Check your work by multiplying out the

factored form.

Factoring Completely

Example Factor completely 6x2 ? 6x ? 36

6x2 ? 6x ? 36 6(x2 ? x ? 6)

Factor out the GCF, 6.

3 terms, the coefficient of x2 is 1.

6(x ? 3)( x 2)

Factor.

3(2) 6 and 3 2 1

Check

6(x ? 3)( x 2) 6(x2 x 6)

6x2 6x 36

?

Factoring Completely

Example Factor completely 1 ? 16x4

1 and 16x4 are both perfect squares so we have

the difference of two squares.

1 ? 16x4

12 ? (4x2)2

(1 ? 4x2)(1 4x2)

Factor. a 1, b 4x2

This is also a difference of two squares.

(1 ? 2x)(1 2x)(1 4x2)

Factor.

Check

(1 ? 2x)(1 2x)(1 4x2) (1 ? 4x2)(1 4x2)

1 ? 16x4

?

Factoring Completely

Example Factor completely 80w3 ? 10

80w3 ? 10 10(8w3 ? 1)

Factor out the GCF, 10.

8w3 and 1 are both perfect cubes, so we have the

difference of two cubes.

10(2w)3 ? 13

a 2w, b 1.

10(2w ? 1)(4w2 2w 1)

Factor.

Check

10(2w ? 1)(4w2 2w 1) 10(8w3 ? 1)

80w3 ? 10

?

Solving Polynomial Equations by Factoring

- Section 6.6

Zero-Product Property

- A polynomial equation is any equation that

contains a polynomial expression. The degree of

a polynomial equation is the degree of the

polynomial expression in the equation.

3z 7 1

4k2 5 0

8x4 4x3 5 12

Degree 1

Degree 2

Degree 4

The Zero-Product Property If the product of two

factors is zero, then at least one of the factors

is 0. That is, if ab 0, then a 0 or b 0 or

both a and b are 0.

Zero Factor Property

Example Solve (x 5)(x 4) 0

Set each expression equal to 0.

x 5 0 or x 4 0

x 5

x ? 4

The solution set is 4, 5.

Check

(x 5)(x 4) 0

(x 5)(x 4) 0

(5 5)(5 4) 0

( 4 5)( 4 4) 0

0(9) 0

9(0) 0

0 0

0 0

?

?

Quadratic Equations

A quadratic equation is an equation equivalent to

one of the form ax2 bx c 0 where

a, b, and c are real numbers and a ? 0.

Steps for Solving a Quadratic Equation by

Factoring Step 1 Write the equation in standard

form, ax2 bx c 0. Step 2 Factor the

expression on the left hand side of the

equation. Step 3 Set each factor found in Step

2 equal to zero using the Zero-Product

Property Step 4 Solve each first-degree

equation for the variable. Step 5 Check Check

your answers by substituting into the original

equation.

Solving Quadratic Equations

Example Solve 6x 18x2 0

6x 18x2 0

Put the equation in standard form.

18x2 6x 0

Factor.

6x(3x 1) 0

Set each factor equal to zero.

6x 0 3x 1 0

x 0

3x ?1

Check

6x 18x2 0

6(0) 18(0)2 0

0 0

?

?

Solving Quadratic Equations

Example Solve 9x2 81

9x2 81 0

Rewrite the equation in standard form.

9(x2 9) 0

Factor out the common factor, 9.

9(x 3)(x 3) 0

Factor the quadratic equation.

9 0 x 3 0 x 3 0

Set each expression equal to 0.

x 3

Solve each equation.

x ? 3

The solution set is 3, 3.

Check

9x2 81

9x2 81

9(? 3)2 81

9(3)2 81

9(9) 81

9(9) 81

?

?

Solving Quadratic Equations

Example Solve

Multiply each side by the LCD.

Simplify.

Remove parentheses.

Write in standard form.

Factor.

Continued.

Solving Quadratic Equations

Example continued

Set each factor equal to zero.

Solve for x.

Check

?

?

Solving Quadratic Equations

Example Solve 3x3 x2 14x

3x3 x2 14x 0

Rewrite the equation in standard form.

x(3x2 x 14) 0

Factor out the common factor, x.

x(x 2)(3x 7) 0

Factor the quadratic equation.

x 0 x 2 0 3x 7 0

Set each expression equal to 0.

Solve each equation.

x 0

x 2

Modeling and Solving Problems with Quadratic

Equations

- Section 6.7

Problems Involving Quadratic Equations

- Example
- The area of a rectangle is 84 square inches.

Determine the length and width if the length is 2

inches less than twice the width.

Step 1. Identify This is a geometry problem

involving the area of a rectangle.

2w 2

l

Step 2. Name

Area 84 l w

w

Let w width

Let l length

Step 3. Translate Because the length is 2

inches less than twice the width, we know that l

2w 2.

84 l w

84 (2w 2)w

Substitute.

Continued.

Problems Involving Quadratic Equations

Example continued

Step 4 Solve

84 (2w 2)w

84 2w2 2w

Distribute.

2w2 2w 84 0

Rewrite the equation in standard form.

2(w2 w 42) 0

Factor out the common factor, 2.

2(w 7)(w 6) 0

Factor the quadratic equation.

2 0 w 7 0 w 6 0

Set each expression equal to 0.

w 7 w 6

Solve the equations.

Continued.

Problems Involving Quadratic Equations

Example continued

The width of the rectangle is 7 inches.

The length is 2w 2 2(7) 2 12 inches.

Step 5 Check.

Area l w

7 12

84

?

The length is 2 inches less than twice the width.

12 2(7) 2

12 12

?

The Pythagorean Theorem

A right triangle is one that contains a right

angle, that is, an angle of 90. The side of the

angle opposite the 90 angle is called the

hypotenuse the remaining two sides are called

legs.

Pythagorean Theorem In a right triangle, the

square of the length of the hypotenuse is equal

to the sum of the squares of the lengths of the

legs. c2 a2 b2 or leg2 leg2

hypotenuse2

The Pythagorean Theorem

- Example
- Find the length of each leg of the right triangle.

The Pythagorean Theorem is used. The lengths of

the legs are 4x and 7x 1, and the length of the

hypotenuse is 9x 1.

a2 b2 c2

(4x)2 (7x 1)2 (9x 1)2

16x2 49x2 14x 1 81x2 18x 1

Continued.

The Pythagorean Theorem

Example continued

16x2 49x2 14x 1 81x2 18x 1

16x2 32x 0

Write in standard form.

16x(x 2) 0

Factor out the GCF.

16x 0

Set each factor equal to 0.

x 2 0

x 0

Solve each equation.

x 2

The length of one leg is 4(2) 8.

The length of the other leg is 7(2) 1 15.

The length of the hypotenuse is 9(2) 1 17.

Check 82 152 172

64 225 289

?

The Pythagorean Theorem

- Example
- A baseball diamond is square. Each side of the

square is 90 feet long. How far is it from home

plate to second base?

Step 1 Identify We want to know how far it is

from home plate to second base.

Home plate

Let c be the distance from home plate to second

base.

Step 2 Name

c

Second base

Continued.

The Pythagorean Theorem

Example continued

Step 3 Translate

c2 a2 b2

Use the Pythagorean Theorem.

c2 902 902

Substitute.

c2 8100 8100

Step 4 Solve

c2 16200

Continued.

The Pythagorean Theorem

Example continued

?127.3 is not used because length is never

negative.

Step 5 Check

c2 a2 b2

127.32 902 902

16205.29 8100 8100

16205.29 ? 16200

?

Step 6 Answer The distance from home plate to

second base is approximately 127.3 feet.

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