Adding and Subtracting Polynomials -...

Adding and Subtracting

Polynomials

Section 9-1

Vocabulary

• Monomial

• Degree of a Monomial

• Polynomial

• Standard Form of a Polynomial

• Degree of a Polynomial

• Binomial

• Trinomial

A monomial is a number, a variable, or a product of numbers and

variables with whole-number exponents.

The degree of a monomial is the sum of the exponents of the

variables. A constant has degree 0.

Definitions

Find the degree of each monomial.

A. 4p4q3

The degree is 7. Add the exponents of the variables:

4 + 3 = 7.

B. 7ed

1+ 1 = 2. C. 3

0 = 0.

Example: Degree of a

Monomial

Find the degree of each monomial.

a. 1.5k2m

2 + 1 = 3.

1 = 1.

b. 2c3

3 = 3.

Your Turn:

You can add or subtract monomials by adding or subtracting like terms.

4a3b2 + 3a2b3 – 2a3b2

Like terms

Not like terms

The variables have the same powers.

The variables have different powers.

Add or subtract like terms by adding or subtracting the coefficients of the like terms.

Like Terms

Identify the like terms in each polynomial.

A. 5x3 + y2 + 2 – 6y2 + 4x3

B. 3a3b2 + 3a2b3 + 2a3b2 – a3b2

Identify like terms. 5x + y + 2 – 6y + 4x 3 2 2 3

Like terms: 5x3 and 4x3, y2 and –6y2

3a b + 3a b + 2a b – a b 3 2 2 3 3 3 2 2 Identify like terms.

Like terms: 3a3b2, 2a3b2, and –a3b2

Example: Identify Like Terms

Identify the like terms in the polynomial.

C. 7p3q2 + 7p2q3 + 7pq2

Identify like terms.

There are no like terms.

7p3q2 + 7p2q3 + 7pq2

Example: Identify Like Terms

Identify the like terms in each polynomial.

A. 4y4 + y2 + 2 – 8y2 + 2y4

B. 7n4r2 + 3n2r3 + 5n4r2 + n4r2

Identify like terms. 4y + y + 2 – 8y + 2y 4 2 2 4

Like terms: 4y4 and 2y4, y2 and –8y2

7n4r2 + 3n2r3 + 5n4r2 + n4r2 Identify like terms.

Like terms: 7n4r2, 5n4r2, and n4r2

Your Turn:

Identify the like terms in the polynomial.

C. 9m3n2 + 7m2n3 + pq2

There are no like terms.

9m3n2 + 7m2n3 + pq2

Your Turn:

Simplify.

A. 4x2 + 2x2

2 6x Combine coefficients:

4 + 2 = 6

Identify like terms. 4x2 + 2x2

Example: Add or Subtract

Monomials

Simplify.

B. 3n5m4 - n5m4

Identify like terms. 3n5m4 - n5m4

Combine coefficients: 3 - 1 = 2. 2n5m4

Example: Add or Subtract

Monomials

Simplify.

A. 2x3 - 5x3

Combine coefficients: 2 - 5 = -3

2x3 - 5x3

Your Turn:

Simplify.

B. 2n5p4 + n5p4

Combine coefficients: 2 + 1 = 3

2n5p4 + n5p4

Your Turn:

A polynomial is a monomial or a sum or difference of

monomials.

Example: 3x4 + 5x2 – 7x + 1

This polynomial is the sum of the

monomials 3x4, 5x2, -7x, and 1.

The degree of a polynomial is the degree of the term

with the greatest degree.

Example: The degree of 3x4 + 5x2 – 7x + 1 is 4.

Definitions

Find the degree of each polynomial.

A. 11x7 + 3x3

11x7: degree 7 3x3: degree 3

The degree of the polynomial is the

greatest degree, 7.

Find the degree of

each term.

Find the degree of

each term.

The degree of the polynomial is the greatest degree, 4.

:degree 3 :degree 4

–5: degree 0

Example: Degree of a

Polynomial

Find the degree of each polynomial.

a. 5x – 6

5x: degree 1 Find the degree of

each term. The degree of the polynomial is the

greatest degree, 1.

b. x3y2 + x2y3 – x4 + 2

x3y2: degree 5

The degree of the polynomial is the

greatest degree, 5.

Find the degree of

each term.

–6: degree 0

x2y3: degree 5

–x4: degree 4 2: degree 0

Your Turn:

The terms of a polynomial may be written in any order.

However, polynomials that contain only one variable

are usually written in standard form.

The standard form of a polynomial that contains one

variable is written with the terms in order from greatest

degree to least degree. When written in standard form,

the coefficient of the first term is called the leading

coefficient.

Example: 3x4 + 5x2 – 7x + 1 and 3 is the leading

coefficient.

Definitions

Write the polynomial in standard form. Then give

the leading coefficient.

6x – 7x5 + 4x2 + 9

Find the degree of each term. Then arrange them in descending order:

6x – 7x5 + 4x2 + 9 –7x5 + 4x2 + 6x + 9

Degree 1 5 2 0 5 2 1 0

–7x5 + 4x2 + 6x + 9. The standard form is The leading

coefficient is –7.

Example: Standard Form

Write the polynomial in standard form. Then give the

leading coefficient.

16 – 4x2 + x5 + 9x3

16 – 4x2 + x5 + 9x3 x5 + 9x3 – 4x2 + 16

Degree 0 2 5 3 0 2 3 5

The standard form is The leading

coefficient is 1.

x5 + 9x3 – 4x2 + 16.

Your Turn:

Write the polynomial in standard form. Then give the

leading coefficient.

18y5 – 3y8 + 14y

18y5 – 3y8 + 14y –3y8 + 18y5 + 14y

Degree 5 8 1 8 5 1

The standard form is The leading

coefficient is –3.

–3y8 + 18y5 + 14y.

Your Turn:

Some polynomials have special names based on

their degree and the number of terms they have.

Degree Name

Constant

Linear

Quadratic

6 or more 6th,7th,degree and so on

Quartic

Quintic

Name Terms

Monomial

Binomial

Trinomial

Polynomial 4 or more

By Degree

By # of Terms

Classify each polynomial according to its degree and

number of terms.

A. 5n3 + 4n

Degree 3 Terms 2

5n3 + 4n is a cubic binomial.

B. 4y6 – 5y3 + 2y – 9

Degree 6 Terms 4

4y6 – 5y3 + 2y – 9 is a

6th-degree polynomial.

C. –2x

Degree 1 Terms 1

–2x is a linear monomial.

Example: Classifying

Polynomials

Classify each polynomial according to its degree and number of

terms.

a. x3 + x2 – x + 2

Degree 3 Terms 4

x3 + x2 – x + 2 is a cubic

polynomial.

Degree 0 Terms 1 6 is a constant monomial.

c. –3y8 + 18y5 + 14y

Degree 8 Terms 3

–3y8 + 18y5 + 14y is an 8th-

degree trinomial.

Your Turn:

Just as you can perform operations on numbers, you can perform operations on polynomials. To add or subtract polynomials, combine like terms.

Adding and Subtracting

Polynomials

Combine like terms.

A. 12p3 + 11p2 + 8p3

12p3 + 11p2 + 8p3

12p3 + 8p3 + 11p2

20p3 + 11p2

Rearrange terms so that like

terms are together. Combine like terms.

B. 5x2 – 6 – 3x + 8

5x2 – 6 – 3x + 8

5x2 – 3x + 8 – 6

5x2 – 3x + 2

Example: Simplifying

Polynomials

Combine like terms.

C. t2 + 2s2 – 4t2 – s2

t2 – 4t2 + 2s2 – s2

t2 + 2s2 – 4t2 – s2

–3t2 + s2

D. 10m2n + 4m2n – 8m2n

10m2n + 4m2n – 8m2n

Combine like terms.

Example: Simplifying

Polynomials

Like terms are constants or terms with the same variable(s) raised to the same power(s).

Remember!

a. 2s2 + 3s2 + s

Combine like terms.

2s2 + 3s2 + s

5s2 + s

b. 4z4 – 8 + 16z4 + 2

4z4 – 8 + 16z4 + 2

4z4 + 16z4 – 8 + 2

20z4 – 6

Combine like terms.

Your Turn:

c. 2x8 + 7y8 – x8 – y8

Combine like terms.

2x8 + 7y8 – x8 – y8

2x8 – x8 + 7y8 – y8

x8 + 6y8

d. 9b3c2 + 5b3c2 – 13b3c2

9b3c2 + 5b3c2 – 13b3c2

Combine like terms.

Your Turn:

Polynomials can be added in either vertical or horizontal form.

In vertical form, align the like terms and add:

In horizontal form, use the Associative and Commutative Properties to regroup and combine like terms.

(5x2 + 4x + 1) + (2x2 + 5x + 2)

= (5x2 + 2x2 + 1) + (4x + 5x) + (1 + 2)

= 7x2 + 9x + 3

5x2 + 4x + 1

+ 2x2 + 5x + 2

7x2 + 9x + 3

Adding Polynomials

A. (4m2 + 5) + (m2 – m + 6)

(4m2 + 5) + (m2 – m + 6)

(4m2 + m2) + (–m) +(5 + 6)

5m2 – m + 11

Group like terms

together. Combine like terms.

B. (10xy + x) + (–3xy + y)

(10xy + x) + (–3xy + y)

(10xy – 3xy) + x + y

7xy + x + y

Group like terms

together. Combine like terms.

Example: Adding

Polynomials

(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)

Group like terms together

within each polynomial.

Combine like terms.

(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)

(6x2 + 3x2 – 8x2) + (3y – 4y – 2y)

Use the vertical method. 6x2 – 4y

+ –5x2 + y

x2 – 3y Simplify.

Example: Adding

Polynomials

Group like terms

together.

Combine like terms.

Your Turn:

Add (5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a).

(5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a)

(5a3 + 7a3) + (3a2 + 12a2) + (–10a – 6a)

12a3 + 15a2 – 16a

Group like terms

together.

Combine like terms.

Your Turn:

To subtract polynomials, remember that subtracting is the same as adding the opposite (distributing the negative). To find the opposite of a polynomial, you must write the opposite of each term in the polynomial:

–(2x3 – 3x + 7)= –2x3 + 3x – 7

Subtracting Polynomials

Subtract.

(x3 + 4y) – (2x3)

(x3 + 4y) + (–2x3)

(x3 – 2x3) + 4y

–x3 + 4y

Rewrite subtraction as addition of

the opposite.

Group like terms together.

Combine like terms.

Example: Subtracting

Polynomials

Subtract.

(7m4 – 2m2) – (5m4 – 5m2 + 8)

(7m4 – 2m2) + (–5m4 + 5m2 – 8)

(7m4 – 5m4) + (–2m2 + 5m2) – 8

(7m4 – 2m2) + (–5m4 + 5m2 – 8)

2m4 + 3m2 – 8

Rewrite subtraction as

addition of the opposite.

Group like terms together.

Combine like terms.

Polynomials

Subtract.

(–10x2 – 3x + 7) – (x2 – 9)

(–10x2 – 3x + 7) + (–x2 + 9)

–10x2 – 3x + 7

–x2 + 0x + 9

–11x2 – 3x + 16

Use the vertical method.

Write 0x as a placeholder.

Combine like terms.

Polynomials

Subtract.

(9q2 – 3q) – (q2 – 5)

(9q2 – 3q) + (–q2 + 5)

9q2 – 3q + 0 + − q2 – 0q + 5

8q2 – 3q + 5

Use the vertical method.

Write 0 and 0q as

placeholders.

Combine like terms.

Your Turn:

Subtract.

(2x2 – 3x2 + 1) – (x2 + x + 1)

(2x2 – 3x2 + 1) + (–x2 – x – 1)

–x2 + 0x + 1 + –x2 – x – 1

–2x2 – x

Use the vertical method. Write 0x as a placeholder.

Combine like terms.

Your Turn:

Practice

• 9-1 Exercises Pg. 459-460 #2-50 even

Multiplying and Factoring

Section 9-2

To multiply monomials and polynomials, you will use some of the properties of exponents that you learned earlier in this chapter.

Multiplying Polynomials

Multiply.

A. (6y3)(3y5)

(6y3)(3y5)

Group factors with like bases together.

B. (3mn2) (9m2n)

(3mn2)(9m2n)

27m3n3

Multiply.

(6 • 3)(y3 • y5)

(3 • 9)(m • m2)(n2 • n)

Multiplying Monomials

Multiply.

Group factors with like bases

together.

Multiply.

(4s2t2)(st)(-12st2)

(4 • -12)(s2 • s • s)(t2 • t • t2)

-48s4t5

When multiplying powers with the same base, keep the base and add the exponents.

x2 x3 = x2+3 = x5

Remember!

Multiply.

a. (3x3)(6x2)

(3x3)(6x2)

(3 • 6)(x3 • x2)

Multiply.

b. (2r2t)(5t3)

(2r2t)(5t3)

(2 • 5)(r2)(t3 • t)

10r2t4

Your Turn:

Multiply.

Group factors with like

bases together.

Multiply.

Your Turn:

(3x2y)(2x3z2)(y4z5)

(3 • 2)(x2 • x3)(y • y4)(z2 • z5)

6x5y5z7

To multiply a polynomial by a monomial, use the Distributive Property.

and Polynomials

Multiply.

4(3x2 + 4x – 8)

(4)3x2 +(4)4x – (4)8

12x2 + 16x – 32

Distribute 4.

Multiply.

Example: Multiplying a

Polynomial by a Monomial

6pq(2p – q)

(6pq)(2p – q)

Multiply.

(6pq)2p + (6pq)(–q)

(6 2)(p p)(q) + (–1)(6)(p)(q q)

12p2q – 6pq2

Distribute 6pq.

Group like bases together.

Multiply.

Example: Multiplying a

Polynomial by a Monomial

Multiply.

a. 2(4x2 + x + 3)

2(4x2 + x + 3)

2(4x2) + 2(x) + 2(3)

8x2 + 2x + 6

Distribute 2.

Multiply.

Your Turn:

Multiply.

b. 3ab(5a2 + b)

3ab(5a2 + b)

(3ab)(5a2) + (3ab)(b)

(3 5)(a a2)(b) + (3)(a)(b b)

15a3b + 3ab2

Distribute 3ab.

Group like bases

together.

Multiply.

Your Turn:

Multiply.

c. 5r2s2(r – 3s)

5r2s2(r – 3s)

(5r2s2)(r) – (5r2s2)(3s)

(5)(r2 r)(s2) – (5 3)(r2)(s2 s)

5r3s2 – 15r2s3

Distribute 5r2s2.

Group like bases

together.

Multiply.

Your Turn:

When multiplying a polynomial by a negative

monomial, be sure to distribute the negative sign.

Helpful Hint

Multiply.

d. –5y3(y2 + 6y – 8)

–5y3(y2 + 6y – 8)

–5y5 – 30y4 + 40y3

Multiply each term in

parentheses by –5y3.

Your Turn:

Factoring

• Factoring a polynomial reverses the

multiplication process (factoring is

unmultiplying).

• When factoring a monomial from a

polynomial, the first step is to find the

greatest common factor (GCF) of the

polynomial’s terms.

Factors that are shared by two or more whole numbers are

called common factors. The greatest of these common factors is

called the greatest common factor, or GCF.

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 32: 1, 2, 4, 8, 16, 32

Common factors: 1, 2, 4

The greatest of the common factors is 4.

Greatest Common Factor

Find the GCF of each pair of numbers.

100 and 60

factors of 100: 1, 2, 4,

5, 10, 20, 25, 50, 100

factors of 60: 1, 2, 3, 4, 5,

6, 10, 12, 15, 20, 30, 60

The GCF of 100 and 60 is 20.

List all the factors.

Circle the GCF.

List the factors.

Example: GCF of Two Numbers

Find the GCF of each pair of numbers.

12 and 16

factors of 12: 1, 2, 3, 4, 6, 12

factors of 16: 1, 2, 4, 8, 16

The GCF of 12 and 16 is 4.

List all the factors.

Circle the GCF.

List the factors.

Your Turn:

You can also find the GCF of monomials that

include variables. To find the GCF of monomials,

write the prime factorization of each coefficient and

write all powers of variables as products. Then find

the product of the common factors.

GCF of Monomials

Find the GCF of each pair of monomials.

15x3 and 9x2

15x3 = 3 5 x x x

9x2 = 3 3 x x

3 x x = 3x2

Write the factorization of each

coefficient and write powers as

products.

Align the common factors.

Find the product of the common

factors.

The GCF of 3x3 and 6x2 is 3x2.

Example: GCF of a

Monomial

8x2 and 7y3

8x2 = 2 2 2 x x

7y3 = 7 y y y

Write the factorization of

each coefficient and write

powers as products.

There are no common

factors other than 1. The GCF 8x2 and 7y is 1.

Example: GCF of a

Monomial

If two terms contain the same variable raised to

different powers, the GCF will contain that variable

raised to the lower power.

Helpful Hint

18g2 and 27g3

18g2 = 2 3 3 g g

27g3 = 3 3 3 g g g

3 3 g g

The GCF of 18g2 and 27g3 is 9g2.

Write the factorization of each

products.

Find the product of the common

factors.

Your Turn:

16a6 and 9b

9b = 3 3 b

16a6 = 2 2 2 2 a a a a a a

Write the

factorization of

each coefficient

and write powers

as products.

Align the common

factors.

There are no common factors

other than 1.

The GCF of 16a6 and 7b is 1.

Your Turn:

8x and 7x2

8x = 2 2 2 x

7v2 = 7 x x

Write the prime factorization of each

products.

The GCF of 8x and 7x2 is x.

Your Turn:

Recall that the Distributive Property states that ab + ac =a(b + c).

The Distributive Property allows you to “factor” out the GCF of

the terms in a polynomial to write a factored form of the

polynomial.

A polynomial is in its factored form when it is written as a

product of monomials and polynomials that cannot be factored

further. The polynomial 2(3x – 4x) is not fully factored because

the terms in the parentheses have a common factor of x.

Factoring out a Monomial

Factor each polynomial. Check your answer.

2x2 – 4

2x2 = 2 x x

4 = 2 2

Find the GCF.

The GCF of 2x2 and 4 is 2.

Write terms as products using the GCF as a

factor. 2x2 – (2 2)

2(x2 – 2)

Check 2(x2 – 2)

2x2 – 4

Multiply to check your answer.

The product is the original polynomial.

Use the Distributive Property to factor out the

Example: Factoring the GCF

Aligning common factors can help you find the greatest

common factor of two or more terms.

Writing Math

8x3 – 4x2 – 16x

2x2(4x) – x(4x) – 4(4x)

4x(2x2 – x – 4)

8x3 – 4x2 – 16x

8x3 = 2 2 2 x x x

4x2 = 2 2 x x

16x = 2 2 2 2 x

2 2 x = 4x

Find the GCF.

The GCF of 8x3, 4x2, and 16x is 4x.

Write terms as products using the

GCF as a factor.

Use the Distributive Property to factor

out the GCF.

The product is the original polynomials.

Factor each polynomial.

–14x – 12x2

– 1(14x + 12x2) Both coefficients are negative.

Factor out –1.

Find the GCF.

The GCF of 14x and 12x2 is 2x.

–1[7(2x) + 6x(2x)]

–1[2x(7 + 6x)]

–2x(7 + 6x)

Write each term as a product using

the GCF.

Use the Distributive Property to

factor out the GCF.

14x = 2 7 x

12x2 = 2 2 3 x x

2 x = 2x

When you factor out –1 as the first step, be sure to include it

in all the other steps as well.

Caution!

3x3 + 2x2 – 10

10 = 2 5

Find the GCF.

other than 1.

The polynomial cannot be factored further.

3x3 + 2x2 – 10

3x3 = 3 x x x

2x2 = 2 x x

5b + 9b3

5b = 5 b

9b = 3 3 b b b

5(b) + 9b2(b)

b(5 + 9b2)

b(5 + 9b2) Check

5b + 9b3

Find the GCF.

The GCF of 5b and 9b3 is b.

The product is the original

polynomial.

GCF as a factor.

out the GCF.

Your Turn:

9d2 – 82

Find the GCF.

other than 1.

The polynomial cannot be factored further.

9d2 – 82

9d2 = 3 3 d d

82 = 2 2 2 2 2 2

Your Turn:

–18y3 – 7y2

– 1(18y3 + 7y2) Both coefficients are negative.

Factor out –1.

Find the GCF.

The GCF of 18y3 and 7y2 is y2.

18y3 = 2 3 3 y y y

7y2 = 7 y y

y y = y2

Write each term as a product using

the GCF.

Use the Distributive Property to

factor out the GCF..

–1[18y(y2) + 7(y2)]

–1[y2(18y + 7)]

–y2(18y + 7)

8x4 + 4x3 – 2x2

8x4 = 2 2 2 x x x x

4x3 = 2 2 x x x

2x2 = 2 x x

2 x x = 2x2

4x2(2x2) + 2x(2x2) –1(2x2)

2x2(4x2 + 2x – 1)

Check 2x2(4x2 + 2x – 1)

8x4 + 4x3 – 2x2

The GCF of 8x4, 4x3 and –2x2 is 2x2.

The product is the original polynomial.

GCF as a factor.

out the GCF.

Find the GCF.

Your Turn:

To write expressions for the length and width of a rectangle

with area expressed by a polynomial, you need to write the

polynomial as a product. You can write a polynomial as a

product by factoring it.

The area of a court for the game squash is 9x2 + 6x m2.

Factor this polynomial to find possible expressions for

the dimensions of the squash court.

A = 9x2 + 6x

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

= 3x(3x + 2)

Possible expressions for the dimensions of the squash court are

3x m and (3x + 2) m.

The GCF of 9x2 and 6x is 3x.

Write each term as a product using the

GCF as a factor.

out the GCF.

Example: Application

What if…? The area of the solar panel on another

calculator is (2x2 + 4x) cm2. Factor this polynomial

to find possible expressions for the dimensions of the

solar panel.

A = 2x2 + 4x

= x(2x) + 2(2x)

= 2x(x + 2)

The GCF of 2x2 and 4x is 2x.

Write each term as a product using the

GCF as a factor.

out the GCF.

Possible expressions for the dimensions of the solar panel are 2x cm,

and (x + 2) cm.

Your Turn:

Assignment

• 9-2 Exercises Pg. 463-464 #1-12 eoo, 13-41

Multiplying Binomials

Section 9-3

Multiplying Polynomials

3 Methods for multiplying polynomials

1. Using the Distributive Property

• Can be used to multiply any two polynomials

2. Using a Table or The Box Method

• Can be used to multiply any two polynomials

3. Using FOIL

• Can only be used to multiply two binomials

To multiply a binomial by a binomial, you can apply the Distributive Property more than once:

(x + 3)(x + 2) = x(x + 2) + 3(x + 2) Distribute x and 3.

Distribute x and 3 again.

Multiply.

Combine like terms.

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

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

= x2 + 2x + 3x + 6

= x2 + 5x + 6

Method 1: Distributive Property

Multiply.

(s + 4)(s – 2)

s(s – 2) + 4(s – 2)

s(s) + s(–2) + 4(s) + 4(–2)

s2 – 2s + 4s – 8

s2 + 2s – 8

Distribute s and 4.

Distribute s and 4 again.

Multiply.

Combine like terms.

Example: Multiply Using

Distributive Property

Multiply.

(a + 3)(a – 4)

a(a – 4)+3(a – 4)

a(a) + a(–4) + 3(a) + 3(–4)

a2 – a – 12

a2 – 4a + 3a – 12

Distribute a and 3.

Distribute a and 3 again.

Multiply.

Combine like terms.

Your Turn:

+ 8 (y – 4)

= (y2 – 4y)

= y2 – 4y + 8y – 32

= y (y – 4)

+ (8y – 32)

= y2 + 4y – 32

Your Turn:

Multiply.

(y + 8)(y – 4)

The product can be simplified using the FOIL

method: multiply the First terms, the Outer terms, the

Inner terms, and the Last terms of the binomials.

2 First Last

Method 2: FOIL

4. Multiply the Last terms. (x + 3)(x + 2) 3 2 = 6

3. Multiply the Inner terms. (x + 3)(x + 2) 3 x = 3x

2. Multiply the Outer terms. (x + 3)(x + 2) x 2 = 2x

(x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6

F O I L

1. Multiply the First terms. (x + 3)(x + 2) x x = x2

Example: Multiply Using FOIL

“First Outer Inner Last”, shortcut for distributing, only

works with binomial-binomial products. Multiply

(x + 3)(x + 2)

= z (z) + z (-12) -6 (z) -6 (-12) F O I L

= z2 - 12z – 6z + 72

= z2 - 18z + 72

= 5x (2x) + 5x (8) -4 (2x) -4 (8) F O I L

= 10x2 + 40x – 8x – 32

= 10x2 + 32x – 32

Example: FOIL

Multiply.

A. (m – 2)(m – 8) B. (x + 3)(x + 4)

(m – 2)(m – 8) (x + 3)(x + 4)

m2 – 8m – 2m + 16 x2 + 4x + 3x + 12

m2 – 10m +16 x2 + 7x +12

Your Turn:

Multiply.

(x – 3)(x – 1)

(x x) + (x(–1)) + (–3 x)+ (–3)(–1) ●

x2 – x – 3x + 3

x2 – 4x + 3

Use the FOIL method.

Multiply.

Combine like terms.

Your Turn:

Multiply.

(2a – b2)(a + 4b2)

2a(a) + 2a(4b2) – b2(a) + (–b2)(4b2)

2a2 + 8ab2 – ab2 – 4b4

2a2 + 7ab2 – 4b4

Use the FOIL method.

Multiply.

Combine like terms.

Your Turn:

To multiply polynomials with more than two terms, you can use the Distributive Property several times. Multiply (5x + 3) by (2x2 + 10x – 6):

(5x + 3)(2x2 + 10x – 6) = 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6)

= 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6)

= 5x(2x2) + 5x(10x) + 5x(–6) + 3(2x2) + 3(10x) + 3(–6)

= 10x3 + 50x2 – 30x + 6x2 + 30x – 18

= 10x3 + 56x2 – 18

Multiply.

(x – 5)(x2 + 4x – 6)

(x – 5 )(x2 + 4x – 6)

x(x2 + 4x – 6) – 5(x2 + 4x – 6)

x(x2) + x(4x) + x(–6) – 5(x2) – 5(4x) – 5(–6)

x3 + 4x2 – 5x2 – 6x – 20x + 30

x3 – x2 – 26x + 30

Distribute x and –5.

Distribute x and −5

again.

Simplify.

Combine like terms.

Example:

Multiply.

(x + 3)(x2 – 4x + 6)

(x + 3 )(x2 – 4x + 6)

x(x2 – 4x + 6) + 3(x2 – 4x + 6)

Distribute x and 3.

Distribute x and 3 again.

x(x2) + x(–4x) + x(6) +3(x2) +3(–4x) +3(6)

x3 – 4x2 + 3x2 +6x – 12x + 18

x3 – x2 – 6x + 18

Simplify.

Combine like terms.

Your Turn:

The width of a rectangular prism is 3 feet less than the height, and the length of the prism is 4 feet more than the height.

Write a polynomial that represents the area of the base of the

prism.

Write the formula for the area of a

rectangle.

Substitute h – 3 for w and h + 4

for l.

A = l w

A = (h + 4)(h – 3)

Multiply. A = h2 + 4h – 3h – 12

Combine like terms. A = h2 + h – 12

The area is represented by h2 + h – 12.

Example: Application

The width of a rectangular prism is 3 feet less than the height, and the length of the prism is 4 feet more than the height.

Find the area of the base when the height is 5 ft.

A = h2 + h – 12

A = 52 + 5 – 12

A = 25 + 5 – 12

A = 18

Write the formula for the area the base of

the prism.

Substitute 5 for h.

Simplify.

Combine terms.

The area is 18 square feet.

Your Turn:

Assignment

• 9-3 pg. #1-19 all #20, 30-38 odd, #39

Multiplying Special

There are formulas (shortcuts) that

work for certain polynomial

multiplication problems.

(a + b)2 = a2 + 2ab + b2

(a - b)2 = a2 – 2ab + b2

(a - b)(a + b) = a2 - b2

Being able to use these formulas will help you in the future when you have to factor. If you do not remember the formulas, you can always multiply

using distributive, FOIL, or the box method.

Let’s try one!

1) Multiply: (x + 4)2

You can multiply this by rewriting this as (x + 4)(x + 4)

You can use the following rule as a shortcut:

(a + b)2 = a2 + 2ab + b2

For comparison, I’ll show you both ways.

1) Multiply (x + 4)(x + 4)

First terms:

Outer terms:

Inner terms:

Last terms:

Combine like terms.

x2 +8x + 16

Now let’s do it with the shortcut!

Notice you

have two of

the same

answer?

1) Multiply: (x + 4)2

using (a + b)2 = a2 + 2ab + b2

a is the first term, b is the second term

(x + 4)2

a = x and b = 4

Plug into the formula a2 + 2ab + b2

(x)2 + 2(x)(4) + (4)2

Simplify.

x2 + 8x+ 16

This is the

same answer!

That’s why

the 2 is in

the formula!

2) Multiply: (3x + 2y)2

using (a + b)2 = a2 + 2ab + b2

(3x + 2y)2

a = 3x and b = 2y

Plug into the formula

a2 + 2ab + b2

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

Simplify

9x2 + 12xy +4y2

Multiply (2a + 3)2

1. 4a2 – 9

2. 4a2 + 9

3. 4a2 + 36a + 9

4. 4a2 + 12a + 9

Multiply: (x – 5)2

using (a – b)2 = a2 – 2ab + b2

Everything is the same except the signs!

(x)2 – 2(x)(5) + (5)2

x2 – 10x + 25

4) Multiply: (4x – y)2

(4x)2 – 2(4x)(y) + (y)2

16x2 – 8xy + y2

Multiply (x – y)2

1. x2 + 2xy + y2

2. x2 – 2xy + y2

3. x2 + y2

4. x2 – y2

5) Multiply (x – 3)(x + 3)

First terms:

Outer terms:

Inner terms:

Last terms:

Combine like terms.

x2 – 9

This is called the difference of squares.

Notice the

middle terms

eliminate

each other!

5) Multiply (x – 3)(x + 3) using

(a – b)(a + b) = a2 – b2

You can only use this rule when the binomials

are exactly the same except for the sign.

(x – 3)(x + 3)

a = x and b = 3

(x)2 – (3)2

x2 – 9

6) Multiply: (y – 2)(y + 2)

(y)2 – (2)2

y2 – 4

7) Multiply: (5a + 6b)(5a – 6b)

(5a)2 – (6b)2

25a2 – 36b2

Multiply (4m – 3n)(4m + 3n)

1. 16m2 – 9n2

2. 16m2 + 9n2

3. 16m2 – 24mn - 9n2

4. 16m2 + 24mn + 9n2

Practice!!

• Pg. 477- 478 #2-8 even, #15-20 all, 28-39

even, #44-52 even

• # 43 answer in complete sentences and turn

in for 5 points.

Factoring x2 + bx + c

Section 9-5

Earlier you learned how to multiply two binomials using the

Distributive Property or the FOIL method. In this lesson, you

will learn how to factor a trinomial into two binominals.

Factoring a Trinomial

x2 + bx + c

Notice that when you multiply (x + 2)(x + 5), the constant term

in the trinomial is the product of the constants in the binomials.

(x + 2)(x + 5) = x2 + 7x + 10

Factoring these trinomials is based on

reversing the FOIL process.

To factor a simple trinomial of the form

x 2 + bx + c (leading coefficient is 1),

express the trinomial as the product of

two binomials. For example,

x 2 + 10x + 24 = (x + 4)(x + 6).

x2 + bx + c

Look at the product of (x + a) and (x + b).

(x + a)(x + b) = x2 + ax + bx + ab

The coefficient of the middle term is the sum of a and b. The

constant term is the product of a and b.

= x2 + (a + b)x + ab

x2 + bx + c

Recall by using the FOIL method that

F O I L

(x + 2)(x + 4) = x 2 + 4x + 2x + 8

= x 2 + 6x + 8

To factor x 2 + bx + c into (x + one #)(x + another #), note that b is the sum of the two numbers and c is the product of the two numbers.

So we’ll be looking for 2 numbers whose product is c and whose sum is b.

Note: there are fewer choices for the product, so that’s why we start there first.

x2 + bx + c

Procedure for Factoring Trinomials

of the Form x 2 + bx + c

Factor:

1. List the factors of 20 (two numbers that

multiply to equal the constant):

2. Select the pairs those sum is 12 (the

middle term).

3. Write the two

binomial factors:

4. Check using FOIL:

2 12 20x x

2 10 20

20 1×20 2×10 4×5

x2 + bx + c

Factoring Trinomials x2 + bx + c TIP

When c is positive, its factors have the same

sign. The sign of b tells you whether the

factors are positive or negative. When b is

positive, the factors are positive and when b

is negative, the factors are negative.

2 12 20

2 9 20

x2 + 6x + 5

Factor each trinomial. Check your answer.

(x + )(x + ) b = 6 and c = 5; look for factors of 5

whose sum is 6.

Factors of 5 Sum

1 and 5 6 The factors needed are 1 and 5.

(x + 1)(x + 5)

Check (x + 1)(x + 5) = x2 + 5x + x + 5 Use the FOIL method.

The product is the

original trinomial.

= x2 + 6x + 5

Example: c is Positive

x2 + 6x + 9

(x + )(x + ) b = 6 and c = 9; look for factors of 9

whose sum is 6.

The factors needed are 3 and 3.

(x + 3)(x + 3)

Check (x + 3)(x + 3 ) = x2 + 3x + 3x + 9 Use the FOIL method. The product is the

original trinomial. = x2 + 6x + 9

Factors of 9 Sum

1 and 9 10

3 and 3 6

x2 – 8x + 15

b = –8 and c = 15; look for

factors of 15 whose sum is –8.

The factors needed are –3 and –5 .

Factors of –15 Sum

–1 and –15 –16

–3 and –5 –8

(x – 3)(x – 5)

Check (x – 3)(x – 5 ) = x2 – 5x – 3x + 15 Use the FOIL method. The product is the

original trinomial. = x2 – 8x + 15

(x + )(x + )

x2 + 8x + 12

b = 8 and c = 12; look for factors of

12 whose sum is 8.

The factors needed are 2 and 6 .

Factors of 12 Sum

1 and 12 13 2 and 6 8

(x + 2)(x + 6)

Check (x + 2)(x + 6 ) = x2 + 6x + 2x + 12 Use the FOIL method.

The product is the

original trinomial.

= x2 + 8x + 12

(x + )(x + )

Your Turn:

x2 – 5x + 6

(x + )(x+ ) b = –5 and c = 6; look for

The factors needed are –2 and –3.

Factors of 6 Sum

–1 and –6 –7 –2 and –3 –5

(x – 2)(x – 3)

Check (x – 2)(x – 3) = x2 – 3x – 2x + 6 Use the FOIL method.

The product is the

original trinomial.

= x2 – 5x + 6

Your Turn:

x2 + 13x + 42

b = 13 and c = 42; look for

factors of 42 whose sum is

The factors needed are 6 and 7.

(x + 6)(x + 7)

Check (x + 6)(x + 7) = x2 + 7x + 6x + 42 Use the FOIL method.

The product is the

original trinomial. = x2 + 13x + 42

(x + )(x + )

Factors of 42 Sum

1 and 42 43

6 and 7 13

2 and 21 23

Your Turn:

x2 – 13x + 40

(x + )(x+ )

b = –13 and c = 40; look for factors

of 40 whose sum is –13.

The factors needed are –5 and –8.

(x – 5)(x – 8)

The product is the

original polynomial.

= x2 – 13x + 40

Factors of 40 Sum

–2 and –20 –22

–4 and –10 –14

–5 and –8 –13

Your Turn:

When c is negative, its factors have opposite

signs. The sign of b tells you which factor is

positive and which is negative. The factor

with the greater absolute value has the same

sign as b.

2 9 22

Factoring Trinomials x2 + bx + c TIP

Factor each trinomial.

x2 + x – 20

(x + )(x + ) b = 1 and c = –20; look for

factors of –20 whose sum is

1. The factor with the greater

absolute value is positive.

The factors needed are 5 and

–1 and 20 19

–2 and 10 8

–4 and 5 1

(x – 4)(x + 5)

Example: c is Negative

Factor each trinomial.

x2 – 3x – 18

b = –3 and c = –18; look for

–3. The factor with the

greater absolute value is

negative. Factors of –18 Sum

1 and –18 –17

2 and – 9 – 7

3 and – 6 – 3

The factors needed are 3 and

–6. (x – 6)(x + 3)

(x + )(x + )

Example: c is Negative

If you have trouble remembering the rules for which

factor is positive and which is negative, you can try

all the factor pairs and check their sums.

Helpful Hint

x2 + 2x – 15

(x + )(x + )

–1 and 15 14

–3 and 5 2

(x – 3)(x + 5)

b = 2 and c = –15; look for

factors of –15 whose sum is 2.

The factor with the greater

absolute value is positive.

The factors needed are –3 and

Check (x – 3)(x + 5) = x2 + 5x – 3x – 15 Use the FOIL method.

The product is the

= x2 + 2x – 15

Your Turn:

x2 – 6x + 8

(x + )(x + ) b = –6 and c = 8; look for

The factors needed are –4

and –2.

Factors of 8 Sum

–1 and –6 –7

–2 and –4 –6

(x – 2)(x – 4)

The product is the

= x2 – 6x + 8

Your Turn:

x2 – 8x – 20

(x – 10)(x + 2)

1 and –20 –19

2 and –10 –8

b = –8 and c = –20; look for

–8. The factor with the

greater absolute value is

negative.

The factors needed are –10

and 2.

(x + )(x + )

Check (x – 10)(x + 2) = x2 + 2x – 10x – 20 Use the FOIL method.

The product is the

= x2 – 8x – 20

Your Turn:

Practice

More Practice

Assignment

• Pg 483-484 # 1-29 odd, 43-54 all

Factoring Trinomials of the

𝑎𝑥2 + 𝑏𝑥 + 𝑐

Factoring

• When the quadratic term (first one) has a

coefficient other than 1, factoring becomes

more difficult. These polynomials are in the

form .

• We follow four steps to factor these

polynomials:

2ax bx c

Steps for “Bottoms Up”

1. Multiply the first coefficient and the last term

together.

2. Find the pair of factors that will combine to

produce the middle term.

3. Put each factor of the pair over a [which is

the leading coefficient] and reduce the

fractions.

4. Use “Bottoms Up” to write the factors of the

trinomial.

What is “Bottoms Up”?

• Once you have your two reduced fractions, how do you

write the factors of the trinomial?

• Let’s suppose these are your fractions:

• Do the following:

becomes _____ and becomes ______

and2 3

2 2x 1 3x 2

should be written as2

FHS Polynomials 145

2 x 20 = 40 We need to find a pair of factors of 24 that combine to give us +5

Multiply the first and last coefficients

22x 3x 20

5 8and

2x 5 x 4

Example 1

1 and -40 2 and -20 4 and -10 5 and -8

-1 and 40 -2 and 20 -4 and 10 -5 and 8

5 4and

4 x 6 = 24 We need to find a pair of factors of 24 that combine to give us +5

Multiply the first and last coefficients

26x 5x 4

3 8and

2x 1 3x 4

Example 1

1 and -24 2 and -12 3 and -8 4 and -6

-1 and 24 -2 and 12 -3 and 8 -4 and 6

1 4and

Example 2

Multiply 8 x 15 = 120. Find all pairs of factors of 120.

We need a pair that combines to 26.

4x 3 2x 5

28x 26x 15 1 and 120

2 and 60

3 and 40

4 and 30

5 and 24

6 and 20

8 and 15

10 and 12

6 20and

3 5and

Practice!!

• Pg. 487-488 #1-12 all, #13-27 eoo,

• #44-47 all

Factoring Special Cases

Section 9-7

Perfect Square Trinomial

Factor the polynomial 25x 2 +

20x + 4.

The result is (5x + 2)2, an example of a

binomial squared.

Any trinomial that factors into a single

binomial squared is called a perfect square

trinomial.

Perfect Square Trinomials

A perfect square trinomial results after

squaring a binomial

• Example: (2x – 5)2

Multiply it out

using FOIL

+ 25 – 10x – 10x

= 4x2 – 20x + 25

The first and last terms are perfect squares.

The middle term is double the product of the

square roots of the first and last terms.

(2x – 5)2 = (2x – 5) (2x – 5)

(a + b)2 = a 2 + 2ab + b 2

(a – b)2 = a 2 – 2ab + b 2

So if the first and last terms of our polynomial to be

factored can be written as expressions squared, and the

middle term of our polynomial is twice the product of

those two expressions, then we can use these two previous

equations to easily factor the polynomial.

a 2 + 2ab + b 2 = (a + b)2

a 2 – 2ab + b 2 = (a – b)2

So how do we factor a Prefect Square

Trinomial

When you have to factor a perfect square

trinomial, the patterns make it easier

Example: Factor 36x2 + 60x + 25

First you have to recognize that it’s a

perfect square trinomial

Perfect

Square

Perfect

Square

Product

Doubled

Product

6x 5 30x

And so, the trinomial factors as: (6x + 5)2

Example – First verify it is a Perfect Square

Trinomial

1. m2 – 4m + 4

2. 9w2 + 24w + 16

3. 16x2 – 72x + 81

4. 25h2 – 100h + 64

5. a2 + 6a + 9

= (m – 2)2

= (3w + 4)2

= (4x – 9)2

= (5h – 16)(5h – 4)

= (a + 3)2

Not a perfect square trinomial! 5h 8 40h

m 2 2m

3w 4 12w

4x 9 36x

a 3 3a

Your Turn:

1. m2 – 6m + 9

2. 4w2 + 28w + 49

3. 81x2 – 18x + 1

4. 9a2 + 12a + 4

= (m – 3)2

= (2w + 7)2

= (9x – 1)2

= (3a + 2)2

m 3 3m

2w 7 14w

9x 1 9x

3a 2 6a

Review: Perfect-Square Trinomial

Difference of Two Squares

D.O.T.S.

Conjugate Pairs

The following pairs of binomials are called conjugates.

Notice that they all have the same terms, only the sign

between them is different.

(3x + 6) and (3x - 6)

(r - 5) and (r + 5)

(2b - 1) and (2b + 1)

(x2 + 5) and (x2 - 5)

Multiplying Conjugates

When we multiply any conjugate pairs, the middle terms always

cancel and we end up with a binomial.

(3x + 6)(3x - 6)

(r - 5)(r + 5)

(2b - 1)(2b + 1)

= 9x2 - 36

= r2 - 25

= 4b2 - 1

Binomials that look like this are called a Difference of Squares:

9x2 - 36

The first term

is a Perfect

Square!

The second term

is a Perfect

Square!

Only TWO terms (a binomial)

A MINUS

between!

A binomial is the difference of two square if

1.both terms are squares and

2.the signs of the terms are different.

9x 2 – 25y 2

– c 4 + d 4

a b a b a b2 2 ( )( )

Factoring the Difference of Two Squares

A Difference of

Squares!

A Conjugate Pair!

Factor the polynomial x 2 – 9.

The first term is a square and the last term, 9, can be

written as 32. The signs of each term are different, so we

have the difference of two squares

Therefore x 2 – 9 = (x – 3)(x + 3).

Note: You can use FOIL method to verify that the

factorization for the polynomial is accurate.

Example:

a b a b a b2 2 ( )( )

Example: Factor x2 - 64

x2 = x • x 64 = 8 • 8

= (x + 8)(x - 8)

Example: Factor 9t2 - 25

9t2 = 3t • 3t 25 = 5 • 5

= (3t + 5)(3t - 5)

Factor x 2 – 16.

Since this polynomial can be written as x 2 – 42,

x 2 – 16 = (x – 4)(x + 4).

Factor 9x2 – 4.

Since this polynomial can be written as (3x)2 – 22,

9x 2 – 4 = (3x – 2)(3x + 2).

Factor 16x 2 – 9y 2.

Since this polynomial can be written as (4x)2 – (3y)2,

16x 2 – 9y 2 = (4x – 3y)(4x + 3y).

Example:

A Sum of Squares?

A Sum of Squares, like x2 + 64,

can NOT be factored!

It is a PRIME polynomial.

Factor x 8 – y 6.

Since this polynomial can be written as

(x 4)2 – (y 3)2,

x 8 – y 6 = (x 4 – y 3)(x 4 + y 3).

Factor x2 + 4.

Oops, this is the sum of squares, not the difference

of squares, so it can’t be factored. This polynomial is a

prime polynomial.

Example:

Factor 36x 2 – 64.

Remember that you should always factor out any

common factors, if they exist, before you start any other

technique.

Factor out the GCF.

36x 2 – 64 = 4(9x 2 – 16)

Since the polynomial can be written as (3x)2 – (4)2,

(9x 2 – 16) = (3x – 4)(3x + 4).

So our final result is 36x 2 – 64 = 4(3x – 4)(3x + 4).

Example:

Recognize a difference of two squares: the coefficients of

variable terms are perfect squares, powers on variable terms

are even, and constants are perfect squares.

Reading Math

Recognizing D.O.T.S.

Determine whether each binomial is a difference of two squares. If

so, factor. If not, explain.

3p 2 – 9q 4

3q 2 3q 2 3p2 is not a perfect square.

3p 2 – 9q 4 is not the difference of two squares because 3p 2 is

not a perfect square.

Your Turn:

100x 2 – 4y 2

Write the polynomial as

(a + b)(a – b).

a = 10x, b = 2y

The polynomial is a difference

of two squares.

100x 2 – 4y 2

2y 2y 10x 10x

(10x)2 – (2y)2

(10x + 2y)(10x – 2y)

100x2 – 4y2 = (10x + 2y)(10x – 2y)

Your Turn:

x 4 – 25y 6

(a + b)(a – b).

a = x2, b = 5y3

The polynomial is a difference

of two squares.

(x 2)2 – (5y 3)2

(x 2 + 5y 3)(x 2 – 5y 3)

x 4 – 25y 6 = (x 2 + 5y 3)(x 2 – 5y 3)

5y 3 5y 3 x 2 x 2

x 4 – 25y 6

Your Turn:

Determine whether each binomial is a difference of two squares.

If so, factor. If not, explain.

1 – 4x 2

(a + b)(a – b).

a = 1, b = 2x

The polynomial is a

difference of two squares.

(1) – (2x)2

(1 + 2x)(1 – 2x)

1 – 4x 2 = (1 + 2x)(1 – 2x)

2x 2x 1 1

1 – 4x 2

Your Turn:

Determine whether each binomial is a difference of two squares.

If so, factor. If not, explain.

p 8 – 49q 6

(a + b)(a – b).

a = p4, b = 7q3

The polynomial is a

difference of two squares.

(p 4)2 – (7q 3)2

(p 4 + 7q 3)(p 4 – 7q 3)

p 8 – 49q 6 = (p 4 + 7q 3)(p 4 – 7q 3)

7q 3 7q 3 ● p 4 p 4 ●

p 8 – 49q 6

Your Turn:

16x 2 – 4y 5

4x 4x 4y5 is not a perfect square.

16x 2 – 4y 5 is not the difference of two squares because 4y 5 is

not a perfect square.

16x 2 – 4y 5

Your Turn:

Review: D.O.T.S.

Practice

• Pg. 493-474 # 1-43 odd, #55

Factoring by Grouping

Section 9-8

Definition

– Using the distributive property to factor polynomials

with four or more terms.

– terms can be put into groups and then factored---- each

group will have a “like” factor used in regrouping.

• Polynomials with four or more terms like 3xy –

21y + 5x – 35, can sometimes be factored by

grouping terms of the polynomials.

• One method is to group the terms into binomials

that can be factored using the distributive

property.

• Then use the distributive property again with a

binomial as the common factor.

A polynomial can be factored by grouping if all of the

following conditions exist.

1. There are four or more terms.

2. Terms have common factors that can be grouped

together, and

3. There are two common factors that are identical.

Symbols: ax + bx + ay + by = (ax + bx) + (ay + by)

= x(a + b) + y(a + b)

= (x + y)(a + b)

Group, factor

Regroup

Factor by Grouping

Factor each polynomial by grouping. Check your

answer.

6h4 – 4h3 + 12h – 8

(6h4 – 4h3) + (12h – 8)

2h3(3h – 2) + 4(3h – 2)

(3h – 2)(2h3 + 4)

Group terms that have a common number

or variable as a factor.

Factor out the GCF of each group.

(3h – 2) is another common factor.

Factor out (3h – 2).

Example: Factor by

Grouping

Check your answer.

Check (3h – 2)(2h3 + 4) Multiply to check your solution.

3h(2h3) + 3h(4) – 2(2h3) – 2(4)

6h4 + 12h – 4h3 – 8

The product is the original

polynomial. 6h4 – 4h3 + 12h – 8

Example: Continued

Factor each polynomial by grouping.

5y4 – 15y3 + y2 – 3y

(5y4 – 15y3) + (y2 – 3y)

5y3(y – 3) + y(y – 3)

(y – 3)(5y3 + y)

Group terms.

Factor out the GCF of each

group.

(y – 3) is a common factor.

Factor out (y – 3).

Example: Factor by

Grouping

6b3 + 8b2 + 9b + 12

(6b3 + 8b2) + (9b + 12)

2b2(3b + 4) + 3(3b + 4)

(3b + 4)(2b2 + 3)

Group terms.

group.

(3b + 4) is a common factor.

Factor out (3b + 4).

Your Turn:

4r3 + 24r + r2 + 6

(4r3 + 24r) + (r2 + 6)

4r(r2 + 6) + 1(r2 + 6)

(r2 + 6)(4r + 1)

Group terms.

group.

(r2 + 6) is a common factor.

Factor out (r2 + 6).

Your Turn:

If two quantities are opposites, their sum is 0

(Additive Inverses).

(5 – x) + (x – 5)

5 – x + x – 5

– x + x + 5 – 5

Helpful Hint

Recognizing opposite binomials can help you factor

polynomials. The binomials (5 – x) and (x – 5) are

opposites. Notice (5 – x) can be written as –1(x – 5).

–1(x – 5) = (–1)(x) + (–1)(–5)

= –x + 5

= 5 – x

So, (5 – x) = –1(x – 5)

Distributive Property.

Simplify.

Commutative Property of

Addition.

Additive Inverse

Factor 2x3 – 12x2 + 18 – 3x

2x3 – 12x2 + 18 – 3x

(2x3 – 12x2) + (18 – 3x)

2x2(x – 6) + 3(6 – x)

2x2(x – 6) + 3(–1)(x – 6)

2x2(x – 6) – 3(x – 6)

(x – 6)(2x2 – 3)

Group terms.

group.

Simplify. (x – 6) is a common

factor.

Factor out (x – 6).

Write (6 – x) as –1(x – 6).

Example: Factoring with

Opposite Groups

15x2 – 10x3 + 8x – 12

(15x2 – 10x3) + (8x – 12)

5x2(3 – 2x) + 4(2x – 3)

5x2(3 – 2x) + 4(–1)(3 – 2x)

5x2(3 – 2x) – 4(3 – 2x)

(3 – 2x)(5x2 – 4)

Group terms.

group.

Simplify. (3 – 2x) is a

common factor.

Factor out (3 – 2x).

Write (2x – 3) as –1(3 – 2x).

Your Turn:

8y – 8 – x + xy

(8y – 8) + (–x + xy)

8(y – 1)+ (x)(–1 + y)

(y – 1)(8 + x)

8(y – 1)+ (x)(y – 1)

Group terms.

group.

(y – 1) is a common factor.

Factor out (y – 1) .

Your Turn:

1. 2x3 + x2 – 6x – 3

2. 7p4 – 2p3 + 63p – 18

(7p – 2)(p3 + 9)

(2x + 1)(x2 – 3)

Your Turn:

Factoring Procedure

Methods of Factoring

Assignment

• Pg. 499-500 # 2-38 even, #46

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