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Slide 3.3- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dividing Polynomials and the 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

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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).

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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

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Slide 3.3- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR LONG DIVISION Step 1Write the terms in the dividend and the divisor in descending powers of the variable. Step 2Insert terms with zero coefficients in the dividend for any missing powers of the variable. Step 3Divide the first terms in the dividend by the first terms in the divisor to obtain the first term in the quotient.

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Slide 3.3- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR LONG DIVISION Step 4Multiply the divisor by the first term in the quotient, and subtract the product from the dividend. Step 5Treat 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.

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Slide 3.3- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Using Long Division Divide Solution

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Slide 3.3- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Using Long Division Solution continued

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Slide 3.3- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Using Long Division Solution continued The quotient is We can write the result in the form The remainder is

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Slide 3.3- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR SYNTHETIC DIVISION Step 1Arrange the coefficients of F(x) in order of descending powers of x, supplying zero as the coefficient of each missing power. Step 2Replace the divisor x – a with a. Step 3Bring 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.

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Slide 3.3- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR SYNTHETIC DIVISION Step 4Add 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 5Multiply 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.

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Slide 3.3- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Using Synthetic Division Use synthetic division to divide Solution The quotient is with remainder –22. So the result is

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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

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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 polynomial Solution By the Remainder Theorem, F(1) is the remainder. The remainder is –2.

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Slide 3.3- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Using the Remainder Theorem Let Solution One way is to evaluate f (x) when x = –3. Another way is synthetic division. Either method yields a remainder of 6.

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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.

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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 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.

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Slide 3.3- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Using the Factor Theorem Solution continued Since the remainder is 0, To find other zeros solve the depressed equation.

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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

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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.

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Slide 3.3- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Solution Since x = 5 represents 1975, we have Hence, 5 is a zero of F(x) = C(x) – 34, and Petroleum Consumption

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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. Petroleum Consumption

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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. Petroleum Consumption

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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 a 0 ; 2. q is a factor of the leading coefficient a n. If is a polynomial function with integer coefficients (a n ≠ 0, a 0 ≠ 0) and is a rational number in lowest terms that is a zero of F(x), then

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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 of List all possible zeros Solution

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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 The remainder of 0 tells us that (x – 1) is a factor of F(x). The other factor is 2x 2 + 7x + 3. To find the other zeros, solve 2x 2 + 7x + 3 = 0.

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Slide 3.3- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 7 Using the Rational Zeros Test Solution continued The solution set is The rational zeros of F are

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