1 © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved Chapter 3 Polynomial and Rational Functions

OBJECTIVES © 2010 Pearson Education, Inc. All rights reserved 2 Dividing Polynomials SECTION Learn the Division Algorithm. Use synthetic division. Use the Remainder and Factor Theorems.

3 © 2010 Pearson Education, Inc. All rights reserved 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).

4 © 2010 Pearson Education, Inc. All rights reserved 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 Either R(x) is the zero polynomial, or the degree of R(x) is less than the degree of D(x).

5 © 2010 Pearson Education, Inc. All rights reserved A 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.

6 © 2010 Pearson Education, Inc. All rights reserved A PROCEDURE FOR LONG DIVISION Step 3 Divide the first term in the dividend by the first term in the divisor to obtain the first term in the quotient. Step 4 Multiply the divisor by the first term in the quotient and subtract the product from the dividend.

7 © 2010 Pearson Education, Inc. All rights reserved A PROCEDURE FOR LONG DIVISION 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. Step 6 Write the quotient and the remainder.

8 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Using Long Division Solution Step 1 Write terms in descending powers of x. Dividend: x 5 + 4x 3 + 2x Divisor: x 2 – x + 1 Step 2 Insert terms with zero coefficients. x 5 + 0x 4 + 4x 3 + 2x 2 + 0x + 7 Find the quotient and remainder when 2x 2 + x x 3 is divided by x – x.

9 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Using Long Division Solution continued Step 3 Divide the first term in the dividend by the first term in the divisor to obtain the first term in the quotient.

10 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Using Long Division Solution continued Step 4 Multiply the divisor by the first term in the quotient and subtract the product from the dividend.

11 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Using Long Division Solution continued Step 5 Treat the remainder obtained in Step 4 as a new dividend and repeat Steps 3 and 4 until a remainder is obtained that is of lower degree than the divisor.

12 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Using Long Division Solution continued Step 5 continued Step 6 Write the quotient and the remainder. Quotient = x 3 + x 2 + 4x + 5 Remainder = x + 2

13 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 2 Using Long Division Solution Because the dividend does not contain an x 3 term, we use a zero coefficient for the missing term.

14 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 2 Using Long Division Solution continued Quotient Remainder The quotient is x 2 + x – 6, and the remainder is x – 1.

15 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 2 Using Long Division Solution continued We can write this result in the form

16 © 2010 Pearson Education, Inc. All rights reserved A PROCEDURE FOR SYNTHETIC DIVISION Divide a polynomial F(x) by x – a. 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.

17 © 2010 Pearson Education, Inc. All rights reserved A PROCEDURE FOR SYNTHETIC DIVISION 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. Step 4 Add the product obtained in the previous step 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.

18 © 2010 Pearson Education, Inc. All rights reserved A PROCEDURE FOR SYNTHETIC DIVISION 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. Step 6 Repeat Step 5 until a product is added to the constant term. Separate the last number below the line with a short vertical line.

19 © 2010 Pearson Education, Inc. All rights reserved A PROCEDURE FOR SYNTHETIC DIVISION Step 7 The last number below the line is the remainder, and the other numbers, reading from left to right, are the coefficients of the quotient, which has degree one less than the dividend F(x).

20 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 4 Using Synthetic Division Use synthetic division to divide 2x 4 + x 3 − 16x by x + 2. Solution Because x – a = x + 2 = x – (– 2), we have a = –2. Write the coefficients of the dividend in a line, supplying 0 as the coefficient of the missing x-term. Then carry out the steps of synthetic division.

21 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 4 Solution continued quotient: 2x 3 – 3x 2 – 10x + 20 The result is remainder: –22 Using Synthetic Division

22 © 2010 Pearson Education, Inc. All rights reserved THE REMAINDER THEOREM If a polynomial F(x) is divided by x – a, then the remainder R is given by

23 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 5 Using the Remainder Theorem Find the remainder when the polynomial Solution By the Remainder Theorem, F(1) is the remainder. The remainder is –2.

24 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 6 Using the Remainder Theorem Let Solution One way is to evaluate f (x) when x = –3. Another way is to use synthetic division. Either method yields a value of 6.

25 © 2010 Pearson Education, Inc. All rights reserved FACTOR THEOREM A polynomial F(x) has (x – a) as a factor if and only if F(a) = 0.

26 © 2010 Pearson Education, Inc. All rights reserved FACTOR THEOREM If F(x) is a polynomial, then the following problems are equivalent: 1. Factoring F(x) 2. Finding the zeros of the function F(x) defined by the polynomial expression 3. Solving (or finding the roots of) the polynomial equation F(x) = 0

27 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 7 Using the Factor Theorem Given that 2 is a zero of the function solve the polynomial equation 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.

28 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 7 Using the Factor Theorem Solution continued Since the remainder is 0, To find other zeros, solve the depressed equation.

29 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 7 Using the Factor Theorem Solution continued Including the original zero of 2, the solution set is

30 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 8 Petroleum Consumption The petroleum consumption C (in quads) in the United States from 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 Find another year between 1975 and 1995 when the model indicates that 34 quads of petroleum were consumed.

31 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 8 Solution Since x = 5 represents 1975, we have Hence, 5 is a zero of F(x) = C(x) – 34, and Petroleum Consumption

32 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 8 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. Petroleum Consumption

33 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 8 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. Petroleum Consumption