Holt Algebra 2 6-4 Factoring Polynomials 6-4 Factoring Polynomials Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson.

Holt Algebra 2

6-4 Factoring Polynomials6-4 Factoring Polynomials

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 2

6-4 Factoring Polynomials

Warm Up

2. Divide by using synthetic division. (x3 – 3x + 5) ÷ (x + 2)

1. Divide by using long division. (8x3 + 6x2 + 7) ÷ (x + 2)

194; –43. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1.

8x2 – 10x + 20 – 33

x + 2

x2 – 2x + 1 + 3

x + 2

Holt 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 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 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 Algebra 2

6-4 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 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 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 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 Algebra 2

6-4 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 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 Algebra 2

6-4 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 Algebra 2

6-4 Factoring Polynomials

Example 4: Geometry ApplicationThe 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 Algebra 2

6-4 Factoring Polynomials

Use synthetic division to factor the polynomial.

1 6 3 –10 1

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 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)