slide 3.3- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

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


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).


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


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.


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.


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



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



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.


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.


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.


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


THE REMAINDER THEOREM

If a polynomial F(x) is divided by x – a, then the remainder R is given by

R F a .


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.


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


THE FACTOR THEOREM

A polynomial F(x) has (x – a) as a factor if and only if F(a) = 0.


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



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



Solution continued

Including the original zero of 2 the solution set is

3,1

3, 2

.


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,


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


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


5 0.003 0.139 1.621 5.005

0.015 0.62 5.005

0.003 0.124 1.001 0


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


0.003x2 0.124x 1.001 0


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


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.



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.



Solution continued2x2 7x 3 0

2x 1 x 3 0

2x 1 0 or x 3 0

x 1

2 or x 3

The solution set is 1, 1

2, 3

.

The rational zeros of F are 3, 1

2, and 1.

slide 3.3- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

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