slide 3.3- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley
TRANSCRIPT
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
OBJECTIVES
Dividing Polynomials andthe Rational Zeros Test
Learn to divide polynomials.
Learn synthetic division.
Learn the Remainder and Factor Theorems.
Learn to use the Rational Zeros Test.
SECTION 3.3
1
2
3
4
Slide 3.3- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
POLYNOMIAL FACTOR
A polynomial D(x) is a factor of a polynomial
F(x) if there is a polynomial Q(x) such that
F(x) = D(x) • Q(x).
Slide 3.3- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
THE DIVISION ALGORITHM
If a polynomial F(x) is divided by a polynomial
D(x), with D(x) ≠ 0, there are unique
polynomials Q(x) and R(x) such that
F(x) = D(x) • Q(x) + R(x)
Either R(x) is the zero polynomial, or the
degree of R(x) is less than the degree of D(x).
Dividend Divisor Quotient Remainder
Slide 3.3- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROCEDURE FOR LONG DIVISION
Step 1 Write the terms in the dividend and the divisor in descending powers of the variable.
Step 2 Insert terms with zero coefficients in the dividend for any missing powers of the variable.
Step 3 Divide the first terms in the dividend by the first terms in the divisor to obtain the first term in the quotient.
Slide 3.3- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROCEDURE FOR LONG DIVISION
Step 4 Multiply the divisor by the first term in the quotient, and subtract the product from the dividend.
Step 5 Treat the remainder obtained in Step 4 as a new dividend, and repeat Steps 3 and 4. Continue this process until a remainder is obtained that is of lower degree than the divisor.
Slide 3.3- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1 Using Long Division
Divide x4 13x2 x 35 by x2 x 6.
Solution
x2 x 6 x4 0x3 13x2 x 35
x4 x3 6x2
x3 7x2 x 35
x3 x2 6x
6x2 7x 35
x2 x
Slide 3.3- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1 Using Long Division
Solution continued
x2 x 6 x4 0x3 13x2 x 35
x4 x3 6x2
x3 7x2 x 35
x3 x2 6x
6x2 7x 35
6x2 6x 36
x 1
x2 x 6
Slide 3.3- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1 Using Long Division
Solution continued
The quotient is
x4 13x2 x 35
x2 x 6x2 x 6
x 1
x2 x 6.
We can write the result in the form
x2 x 6.
The remainder is x 1.
Slide 3.3- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROCEDURE FOR SYNTHETIC DIVISION
Step 1 Arrange the coefficients of F(x) in order of descending powers of x, supplying zero as the coefficient of each missing power.
Step 2 Replace the divisor x – a with a.Step 3 Bring the first (leftmost) coefficient
down below the line. Multiply it by a, and write the resulting product one column to the right and above the line.
Slide 3.3- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROCEDURE FOR SYNTHETIC DIVISION
Step 4 Add the product obtained in Step 3 to the coefficient directly above it, and write the resulting sum directly below it and below the line. This sum is the “newest” number below the line.
Step 5 Multiply the newest number below the line by a, write the resulting product one column to the right and above the line, and repeat Step 4.
Slide 3.3- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Using Synthetic Division
Use synthetic division to divide2x4 x3 16x2 18 by x 2.
2x3 3x2 10x 20 22
x 2.
Solution
2 2 1 16 0 18
4 6 20 40
2 3 10 20 22
2x3 3x2 10x 20The quotient is with remainder –22. So the result is
Slide 3.3- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
THE REMAINDER THEOREM
If a polynomial F(x) is divided by x – a, then the remainder R is given by
R F a .
Slide 3.3- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3 Using the Remainder Theorem
Find the remainder when the polynomialF x 2x5 4x3 5x2 7x 2
is divided by x 1.
F 1 2 1 5 4 1 3 5 1 2 7 1 2
2 4 5 7 2 2
Solution
By the Remainder Theorem, F(1) is the remainder.
The remainder is –2.
Slide 3.3- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Using the Remainder Theorem
Let f x x4 3x3 5x2 8x 75. Find f 3 .Solution
One way is to evaluate f (x) when x = –3.
Another way is synthetic division.
f 3 3 4 3 3 3 5 3 2 8 3 75 6
Either method yields a remainder of 6.
3 1 3 5 8 75
3 0 15 69
1 0 5 23 6
Slide 3.3- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
THE FACTOR THEOREM
A polynomial F(x) has (x – a) as a factor if and only if F(a) = 0.
Slide 3.3- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5 Using the Factor Theorem
Given that 2 is a zero of the functionf x 3x2 2x2 19x 6, solve the
polynomial equation 3x2 2x2 19x 6 0.
Solution
Since 2 is a zero of f (x), f (2) = 0 and (x – 2) is a factor of f (x). Perform synthetic division by 2.
2 3 2 19 6
6 16 6
3 8 3 0
Slide 3.3- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5 Using the Factor Theorem
f x 3x2 2x2 19x 6 x 2 3x2 8x 3
Solution continued
Since the remainder is 0,
To find other zeros solve the depressed equation.
3x2 8x 3 0
3x 1 x 3 0
3x 1 0 or x 3 0
x 1
3 or x 3
Slide 3.3- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5 Using the Factor Theorem
Solution continued
Including the original zero of 2 the solution set is
3,1
3, 2
.
Slide 3.3- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Petroleum Consumption
The petroleum consumption C (in quads) in the United States from 1970 - 1995 can be modeled by the function
where x = 0 represents 1970, x = 1 represents 1971, and so on.The model indicates that C(5) = 34 quads of petroleum were consumed in 1975. Find another year between 1975 and 1995 when the model indicates that 34 quads of petroleum were consumed.
C x 0.003x3 0.139x2 1.621x 28.995,
Slide 3.3- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6
Solution
Since x = 5 represents 1975, we have
C 5 34
C x x 5 Q x 34
C x 34 x 5 Q x
Hence, 5 is a zero of F(x) = C(x) – 34, and
F x C x 34
0.003x3 0.139x2 1.621x 5.005.
Petroleum Consumption
Slide 3.3- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6
Solution continued
We need to find another zero between 5 and 25. Since 5 is a zero perform synthetic division by 5.
Solve the depressed equation Q(x) = 0.
0.003x2 0.124x 1.001 0
Petroleum Consumption
5 0.003 0.139 1.621 5.005
0.015 0.62 5.005
0.003 0.124 1.001 0
Slide 3.3- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6
Solution continued
Use the quadratic formula.
Since only 11 is between 5 and 25, we use this value. In 1981 (the year corresponding to x = 11) the petroleum consumption was 34 quads.
x 0.124 0.124 2 4 0.003 1.001
2 0.003 x 11 or x 30.3
Petroleum Consumption
0.003x2 0.124x 1.001 0
Slide 3.3- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
THE RATIONAL ZEROS TEST
1. p is a factor of the constant term a0;
2. q is a factor of the leading coefficient an.
If
is a polynomial function with integer
coefficients (an ≠ 0, a0 ≠ 0) and is a
rational number in lowest terms that is a
zero of F(x), then
F x an xn an 1xn 1 ... a2 x2 a1x a0
p
q
Slide 3.3- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Using the Rational Zeros Test
Find all the rational zeros ofF x 2x3 5x2 4x 3.
List all possible zeros
Solution
p
q
Factors of the constant term, 3
Factors of the leading coefficient, 2
Factors of 3 : 1, 3
Factors of 2 : 1, 2
Possible rational zeros are: 1, 1
2,
3
2, 3.
Slide 3.3- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Using the Rational Zeros Test
Begin testing with 1, if it is not a rational zero, then try another possible zero.
Solution continued
1 2 5 4 3
2 7 3
2 7 3 0
The remainder of 0 tells us that (x – 1) is a factor of F(x). The other factor is 2x2 + 7x + 3. To find the other zeros, solve 2x2 + 7x + 3 = 0.