Factoring Polynomials 3-4

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Warm Up

Lesson Presentation

Lesson Quiz

Holt McDougal Algebra 2

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5-Minute Check 11/8

2. a2 – b2

3. (x – 1)(x + 3) 4. (a + 1)(a2 + 1)

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5-Minute Check 11/9

Holt McDougal Algebra 2

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5-Minute Check 11/8

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SW use the Factor Theorem to

determine factors of a polynomial.

SW factor the sum and difference of two

squares and cubes.

Holt McDougal Algebra 2

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_{Factoring Polynomials}

SW use the Factor Theorem to

determine factors of a polynomial.

SW factor the sum and difference of two

squares and cubes.

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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

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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. B. (x + 2); (3x4 + 6x3 – 5x – 10)

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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 –4 5 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 10 0 0 –5 0 0 P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.

Holt McDougal Algebra 2

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Example 2

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

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Example 2

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 –10 25 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 10 4 0 5 2 0 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

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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.

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Factor: x3 – x2 – 25x + 25.

Holt McDougal Algebra 2

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Factor: x3 – x2 – 25x + 25.

Example 3: 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.

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Example 3 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

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Example 4

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Example 4

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.

Holt McDougal Algebra 2

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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. ✔

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Example 5

Holt McDougal Algebra 2

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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)

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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

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Example 6: 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)

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Example 7: 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 a}3 _{– b}3 ₌

(a – b) × (a2 + ab + b2).

Holt McDougal Algebra 2

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Example 8

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 a}3 _{+ b}3 ₌

(a + b) × (a2 – ab + b2).

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Example 9

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

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Example 10: 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

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Use synthetic division to factor the polynomial.

1 6 3 –10 1 1 1 0 V(x)= (x – 1)(x2 + 7x + 10) 10 7 10 7 Write V(x) as a product. V(x)= (x – 1)(x + 2)(x + 5) _{Factor the quadratic.}

Example 10 Continued

Holt McDougal Algebra 2

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Example10

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

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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.}

Example 10 Continued

Holt McDougal Algebra 2

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Factor each expression.

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