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

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

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 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)