Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial...

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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide Chapter 6 Polynomial Functions

Transcript of Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial...

Page 1: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1

Chapter 6Polynomial Functions

Page 2: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 2

6.4 Factoring Trinomials of the Form x2 + bx + c; Factoring Out the GCF

Page 3: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 3

Comparing Multiplying with Factoring

Multiplying and factoring are reverse processes. For example,

Page 4: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 4

Factoring a Trinomial of the Form x2 + bx + c

To see how to factor x2 + 5x + 6, let’s take another look at how we find the product (x + 2)(x + 3):

Page 5: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Example: Factoring a Trinomial of the Form x2 + bx + c

Factor x2 + 11x + 24.

Page 6: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

We need two integers whose product is 24 and whose sum is 11.

Since 3(8) = 24 and 3 + 8 = 11, we conclude that the last terms of the factors are 3 and 8.

Page 7: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

x2 + 11x + 24 = (x + 3)(x + 8)

Check by finding the product of the result:

(x + 3)(x + 8) = x2 + 8x + 3x + 24 = x2 + 11x + 24

By the commutative law, (x + 3)(x + 8) = (x + 8)(x + 3),

so we can write the factors in either order.

Page 8: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 8

Factoring x2 + bx + c

To factor x2 + bx + c, look for two integers p and q whose product is c and whose sum is b. That is pq = c and p + q = b. If such integers exist, the factored polynomial is

(x + p)(x + q)

Page 9: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 9

Factoring x2 + bx + c with c Positive

To factor a trinomial of the form x2 + bx + c with a positive constant term c,

• If b is positive, look for two positive integers whose product is c and whose sum is b. For example,

Page 10: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 10

Factoring x2 + bx + c with c Positive

• If b is negative, look for two negative integers whose product is c and whose sum is b. For example,

Page 11: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 11

Example: Factring a Trinomial of the Form x2 + bx + c

Factor w2 – 3w – 18.

Page 12: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

We need two integers whose product is –18 and whose sum is –3. Since the product is negative, the two integers must have difference signs. Here are the possibilities:

Page 13: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

Since 3(–6) = –18 and 3 + (–6) = –3, we conclude that the last terms of the factors are 3 and –6.

(w + 3)(w – 6)

Check by finding the product:

(w + 3)(w – 6) = w2 – 6w + 3w – 18 = w2 – 3w – 18

Page 14: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 14

Factoring x2 + bx + c with c Negative

To factor a trinomial of the form x2 + bx + c with a negative constant term c, look for two integers with different signs whose product is c and whose sum is b. For example,

Page 15: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 15

Example: Factoring a Trinomial with Two Variables

Factor a2 + 6ab + 8b2.

Page 16: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

Write the trinomial in the form a2 + (6b)a + 8b2. We need two monomials whose product is 8b2 and whose sum is 6b. So, the last two terms are 2b and 4b.

a2 + 6ab + 8b2 = (a + 2b)(a + 4b)

Check by finding the product.

(a + 2b)(a + 4b) = a2 + 6ab + 8b2

Page 17: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Prime Polynomials

Just as a prime number has no positive factors other than itself and 1, a polynomial that cannot be factored is called prime.

Page 18: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Example: Identifying a Prime Polynomial

Factor –14 + 6x + x2.

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Solution

Write the polynomial in descending order:

x2 + 6x – 14

Page 20: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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SolutionWe need two integers whose product is –14 and whose sum is 6. Since the product is negative, the integers must have different signs. Here are the possibilities:

Because none of the sums equal 6, we conclude that the trinomial is prime.

Page 21: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Factoring Out the GCF

Definition

The greatest common factor (GCF) of two or more terms is the monomial with the largest coefficient and the highest degree that is a factor of all the terms.

Page 22: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Example: Factoring Out the GCF

Factor 18x4 – 30x2.

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Solution

Begin by factoring 18x4 and 30x2:

18x4 = 2 ∙ 3 ∙ 3 ∙ x ∙ x ∙ x ∙ x30x2 = 2 ∙ 3 ∙ 5 ∙ x ∙ x

There are four common factors, shown in blue. So, the GCF is 6x2:

2 224 218 30 6 63 5xx x xx 2 236 5x x

Page 24: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

Use a graphing calculator table to verify our work.

Page 25: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Example: Completely Factoring a Polynomial

Factor 3x3 + 21x2 + 36x.

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Solution

The GCF is 3x:

3x3 + 21x2 + 36x = 3x(x2 + 7x + 12)

Page 27: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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SolutionTemporarily put aside the GCF, 3x.

To factor x2 + 7x + 12, we need two integers whose product is 12 and whose sum is 7:

Because 3(4) = 12 and 3 + 4 = 7, we conclude that the last terms of the factors are 3 and 4.

Page 28: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

2 7 12 ( 3 4)( )x x x x

So,

3 2 23 21 36 3 7 12 3 ( 3)( 4)x x x x x x x x x

Page 29: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Factor Out the GCF First

In general, when the leading coefficient of a polynomial is positive and the GCF is not 1, first factor out the GCF.

But, don’t forget to include the GCF as part of your answer!

Page 30: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Factoring Completely

Warning

When factoring a polynomial, always completely factor it.

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Factoring when the Leading Coefficient is Negative

When the leading coefficient of a polynomial is negative, first factor out the opposite of the GCF.

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Example: Factoring Out the Opposite of the GCF

Factor –2r4 + 18r3 – 40r2.

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Solution

For this polynomial, the GCF is 2r2. The leading coefficient of –2r4 + 18r3 – 40r2 is –2, which is negative. So, first factor out the opposite of the GCF:

4 3 2 2 22 18 40 2 9 20r r r r r r 22 ( 5)( 4)r r r

Page 34: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.4, Slide 1 Chapter 6 Polynomial Functions.

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Solution

Use a graphing calculator table to verify our work.

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Summary

1. If the leading coefficient of a polynomial is positive and the GCF is not 1, first factor out the GCF. If the leading coefficient is negative, first factor out the opposite of the GCF.

2. Always completely factor a polynomial.