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Holt McDougal Algebra 2

Factoring PolynomialsFactoring Polynomials

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz

Holt McDougal Algebra 2

Holt McDougal Algebra 2

Factoring Polynomials

Warm UpFactor each expression.

1. 3x – 6y

2. a2 – b2

3. (x – 1)(x + 3)

4. (a + 1)(a2 + 1)

x2 + 2x – 3

3(x – 2y)

(a + b)(a – b)

a3 + a2 + a + 1

Find each product.

Holt McDougal Algebra 2

Factoring Polynomials

Use the Factor Theorem to determine factors of a polynomial.

Factor the sum and difference of two cubes.

Objectives

Holt McDougal Algebra 2

Factoring Polynomials

Recall that if a number is divided by any of its factors, the remainder is 0. Likewise, if a polynomial is divided by any of its factors, the remainder is 0. The Remainder Theorem states that if a polynomial is divided by (x – a), the remainder is the value of the function at a. So, if (x – a) is a factor of P(x), then P(a) = 0.

Holt McDougal Algebra 2

Factoring Polynomials

Determine whether the given binomial is a factor of the polynomial P(x).

Example 1: Determining Whether a Linear Binomial is

a Factor

A. (x + 1); (x2 – 3x + 1)

Find P(–1) by synthetic substitution.

1 –3 1 –1

–1

1 5–4

4

P(–1) = 5

P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1.

B. (x + 2); (3x4 + 6x3 – 5x – 10)

Find P(–2) by synthetic substitution.

3 6 0 –5 –10 –2

–6

3 0

1000

–500

P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.

Holt McDougal Algebra 2

Factoring Polynomials

Check It Out! Example 1

Determine whether the given binomial is a factor of the polynomial P(x).

a. (x + 2); (4x2 – 2x + 5)

Find P(–2) by synthetic substitution.

4 –2 5 –2

–8

4 25–10

20

P(–2) = 25

P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5.

b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30)

1 –2 2 1 –10 2

2

1 0

1040

520

P(2) = 0, so (3x – 6) is a

factor of P(x) = 3x4 – 6x3 +

6x2 + 3x – 30.

Divide the polynomial by 3,

then find P(2) by synthetic

substitution.

Holt McDougal Algebra 2

Factoring Polynomials

You are already familiar with methods for factoring quadratic expressions. You can factor polynomials of higher degrees using many of the same methods you learned in Lesson 5-3.

Holt McDougal Algebra 2

Factoring Polynomials

Factor: x3 – x2 – 25x + 25.

Example 2: Factoring by Grouping

Group terms.(x3 – x2) + (–25x + 25)

Factor common monomials

from each group.x2(x – 1) – 25(x – 1)

Factor out the common

binomial (x – 1).(x – 1)(x2 – 25)

Factor the difference of

squares.(x – 1)(x – 5)(x + 5)

Holt McDougal Algebra 2

Factoring Polynomials

Example 2 Continued

Check Use the table feature of your calculator to compare the original expression and the factored form.

The table shows that the original function and the factored form have the same function values.

Holt McDougal Algebra 2

Factoring Polynomials

Check It Out! Example 2a

Factor: x3 – 2x2 – 9x + 18.

Group terms.(x3 – 2x2) + (–9x + 18)

Factor common monomials

from each group.x2(x – 2) – 9(x – 2)

Factor out the common

binomial (x – 2).(x – 2)(x2 – 9)

Factor the difference of

squares.(x – 2)(x – 3)(x + 3)

Holt McDougal Algebra 2

Factoring Polynomials

Check It Out! Example 2a Continued

Check Use the table feature of your calculator to compare the original expression and the factored form.

The table shows that the original function and the factored form have the same function values.

Holt McDougal Algebra 2

Factoring Polynomials

Check It Out! Example 2b

Factor: 2x3 + x2 + 8x + 4.

Group terms.(2x3 + x2) + (8x + 4)

Factor common monomials

from each group.x2(2x + 1) + 4(2x + 1)

Factor out the common

binomial (2x + 1).(2x + 1)(x2 + 4)

(2x + 1)(x2 + 4)

Holt McDougal Algebra 2

Factoring Polynomials

Just as there is a special rule for factoring the difference of two squares, there are special rules for factoring the sum or difference of two cubes.

Holt McDougal Algebra 2

Factoring Polynomials

Example 3A: Factoring the Sum or Difference of Two

Cubes

Factor the expression.

4x4 + 108x

Factor out the GCF, 4x.4x(x3 + 27)

Rewrite as the sum of cubes.4x(x3 + 33)

Use the rule a3 + b3 = (a + b)

(a2 – ab + b2).4x(x + 3)(x2 – x 3 + 32)

4x(x + 3)(x2 – 3x + 9)

Holt McDougal Algebra 2

Factoring Polynomials

Example 3B: Factoring the Sum or Difference of Two

Cubes

Factor the expression.

125d3 – 8

Rewrite as the difference

of cubes.(5d)3 – 23

(5d – 2)[(5d)2 + 5d 2 + 22] Use the rule a3 – b3 =

(a – b) (a2 + ab + b2).

(5d – 2)(25d2 + 10d + 4)

Holt McDougal Algebra 2

Factoring Polynomials

Check It Out! Example 3a

Factor the expression.

8 + z6

Rewrite as the difference

of cubes.(2)3 + (z2)3

(2 + z2)[(2)2 – 2 z + (z2)2] Use the rule a3 + b3 =

(a + b) (a2 – ab + b2).

(2 + z2)(4 – 2z + z4)

Holt McDougal Algebra 2

Factoring Polynomials

Check It Out! Example 3b

Factor the expression.

2x5 – 16x2

Factor out the GCF, 2x2.2x2(x3 – 8)

Rewrite as the difference of

cubes.2x2(x3 – 23)

Use the rule a3 – b3 = (a – b)

(a2 + ab + b2).2x2(x – 2)(x2 + x 2 + 22)

2x2(x – 2)(x2 + 2x + 4)

Holt McDougal Algebra 2

Factoring Polynomials

Example 4: Geometry Application

The volume of a plastic storage box is modeled by the function V(x) = x3 + 6x2 + 3x – 10. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).

V(x) has three real zeros at x = –5, x = –2, and x = 1. If the model is accurate, the box will have no volume if x = –5, x = –2, or x = 1.

Holt McDougal Algebra 2

Factoring Polynomials

Use synthetic division to

factor the polynomial.

1 6 3 –101

1

1 0

V(x)= (x – 1)(x2 + 7x + 10)

107

107

Write V(x) as a product.

V(x)= (x – 1)(x + 2)(x + 5) Factor the quadratic.

Example 4 Continued

One corresponding factor is (x – 1).

Holt McDougal Algebra 2

Factoring Polynomials

Check It Out! Example 4

The volume of a rectangular prism is modeled by the function V(x) = x3 – 8x2 + 19x – 12, which is graphed below. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).

V(x) has three real zeros at x = 1, x = 3, and x = 4. If the model is accurate, the box will have no volume if x = 1, x = 3, or x = 4.

Holt McDougal Algebra 2

Factoring Polynomials

Use synthetic division to

factor the polynomial.

1 –8 19 –121

1

1 0

V(x)= (x – 1)(x2 – 7x + 12)

12–7

12–7

Write V(x) as a product.

V(x)= (x – 1)(x – 3)(x – 4) Factor the quadratic.

Check It Out! Example 4 Continued

One corresponding factor is (x – 1).

Holt McDougal Algebra 2

Factoring Polynomials

4. x3 + 3x2 – 28x – 60

Lesson Quiz

2. x + 2; P(x) = x3 + 2x2 – x – 2

1. x – 1; P(x) = 3x2 – 2x + 5

8(2p – q)(4p2 + 2pq + q2)

(x + 3)(x + 3)(x – 3)3. x3 + 3x2 – 9x – 27

P(1) ≠ 0, so x – 1 is not a factor of P(x).

P(2) = 0, so x + 2 is a factor of P(x).

4. 64p3 – 8q3

(x + 6)(x – 5)(x + 2)