Holt McDougal Algebra 2

6-4 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

6-4 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

6-4 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

6-4 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

6-4 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

6-4 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

6-4 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

6-4 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

6-4 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

6-4 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)