copyright © 2015, 2008, 2011 pearson education, inc. section 6.3, slide 1 chapter 6 polynomial...

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 1

Chapter 6Polynomial Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 2

6.3 Dividing Polynomials: Long Division and Synthetic Division

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 3

Dividing by a Monomial

If A, B, and C are monomials and B is nonzero, then

In words: To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

A C A CB B B

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 4

Example: Dividing by a Monomial

Find the quotient.

4 3 3 2 4

2

6 8 42

x y x y x yx y

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Solution

2 2

4 3 3

2 2

2 4 4 3 3 2 46 8 4 62 2 2 2

8 4x y x y x y x y xx y x y x y

y xy

yx

2 2 33 4 2x xy y

2 2 2 3 4 3 3 2 42 3 4 2 6 8 4x y x xy y x y x y x y

Verify our work by finding the product:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 6

Using Long Division to Divide by a Binomial

We can use long division to divide a polynomial by a binomial. The steps are similar to performing long division with numbers.

Recall that we can verify our work by checking that

Divisor ∙ Quotient + Remainder = Dividend

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 7

Example: Dividing by a Binomial

Divide.22 11 15

3x x

x

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Solution

Divide 2x2 (the first term of the dividend) by x (the first term of the divisor): 22

.2xx

x

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Solution

To subtract 2x2 + 6x, we change the signs of 2x2 and 6x and add:

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Solution

Next, bring down the 15:

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Solution

The repeat the process.

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Solution

We conclude:

To check:

which is true.

22 11 153

2 5x x

xx

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 13

Example: Dividing a Polynomial with Missing Terms

Divide.

3 2

2

1 5 4 23

x x xx

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 14

Solution

First, write the dividend in decreasing order:

5x3 + 2x2 – 4x + 1

Also, since the divisor doesn’t have an x term, we use 0x as a placeholder:

x2 + 0x – 3

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Solution

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Solution

We conclude:

To check:

which is true.

2

2

3

2 311

34 75 2 1

5 2xxx x x

x x

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 17

Synthetic Division

When we divide a polynomial by a binomial of the form x – a, we can use a method called synthetic division, which uses the ideas of long division but is more efficient.

To perform synthetic division, the divisor must be of the form x – a.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 18

Example: Performing Synthetic Division

Use synthetic division to divide:

3 23 7 10 122

x x xx

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Solution

When performing synthetic division, there is no need to write the coefficient of x of the divisor x – 2, The method already takes the divisions by x into account.

Also, instead of dividing by the constant term –2, we divide by its opposite: 2. That way, we can add without having to change signs first.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 20

Solution

Start by writing 2 (of x – 2) and the coefficients of the dividend. Then bring down the 3.

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Solution

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Solution

So, the result is 3x2 – x + 8 with a remainder 4.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 23

Solution

We conclude:

3 223 7 10 12 4

3 82 2

x x xx x

x x

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 24

Example: Performing Synthetic Division when There Are Missing Terms

Use synthetic division to divide:

32 22 123

p pp

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 25

Solution

Because the dividend does not have a p2 term, we use 0 as a placeholder:

We conclude:3

22 22 122 6 4

3p p

p pp