copyright © 2015, 2008, 2011 pearson education, inc. section 6.3, slide 1 chapter 6 polynomial...
TRANSCRIPT
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 1
Chapter 6Polynomial Functions
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 2
6.3 Dividing Polynomials: Long Division and Synthetic Division
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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
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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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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 5
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:
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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
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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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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 8
Solution
Divide 2x2 (the first term of the dividend) by x (the first term of the divisor): 22
.2xx
x
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 9
Solution
To subtract 2x2 + 6x, we change the signs of 2x2 and 6x and add:
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 10
Solution
Next, bring down the 15:
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 11
Solution
The repeat the process.
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 12
Solution
We conclude:
To check:
which is true.
22 11 153
2 5x x
xx
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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
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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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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 15
Solution
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 16
Solution
We conclude:
To check:
which is true.
2
2
3
2 311
34 75 2 1
5 2xxx x x
x x
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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.
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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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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 19
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.
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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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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 21
Solution
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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 22
Solution
So, the result is 3x2 – x + 8 with a remainder 4.
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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
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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
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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