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

Holt McDougal Algebra 2

6-4 Factoring Polynomials6-4 Factoring Polynomials

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz



6-4 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.



Use the Factor Theorem to determine factors of a polynomial.Factor the sum and difference of two cubes.

Objectives



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.



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.



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.



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.



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)



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.



Check It Out! Example 2a

Factor: x3 – 2x2 – 9x + 18.

Group terms.(x3 – 2x2) + (–9x + 18)


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)



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.



Check It Out! Example 2b

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

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


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

Factor out the common binomial (2x + 1).

(2x + 1)(x2 + 4)

(2x + 1)(x2 + 4)



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.



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)



Example 3B: Factoring the Sum or Difference of Two Cubes


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)



Check It Out! Example 3a


8 + z6


(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)



Check It Out! Example 3b


2x5 – 16x2

Factor out the GCF, 2x2.2x2(x3 – 8)


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)



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.



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).



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.



Use synthetic division to factor the polynomial.

1 –8 19 –12 1

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).



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)

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