Explanation Procedure used: Long Division of polynomials: “1. Arrange the terms of both the dividend and the divisor in descending powers of any variable. 2. Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient. 3. Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms line up. 4. Subtract the product from the dividend. 5. Bring down the next term in the original dividend and write it next to the remainder to form a new dividend. 6. Use this new expression as the dividend and repeat this process until the remainder can no longer be divided. This will occur when the degree of the remainder (the highest exponent on a variable in the remainder) is less than the degree of the divisor”. Calculation: Divide the polynomial 2 x 5 − x 2 + 8 x + 44 by x + 1 as follows. The dividend 2 x 5 − x 2 + 8 x + 44 and the divisor x + 1 are in descending powers of x . x + 1 2 x 5 − x 2 + 8 x + 44 Divide the first term of the dividend which is 2 x 5 by the first term of the divisor which is x . That is, 2 x 5 x = 2 x 4 . Multiply each term in x + 1 by 2 x 4 and then align terms of product under like terms in dividend. That is, 2 x 4 ( x + 1 ) = 2 x 5 + 2 x 4 . Proceed in this similar manner, to divided the polynomial 2 x 5 − x 2 + 8 x + 44 by x + 1 as follows. x + 1 2 x 4 − 2 x 3 + 2 x 2 − 3 x + 11 2 x 5 − x 2 + 8 x + 44 2 x 5 + 2 x 4 _ − 2 x 4 − x 2 − 2 x 4 − 2 x 3 _ 2 x 3 − x 2 + 8 x 2 x 3

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 15.1, Problem 7E

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To divide: The polynomial 2x5−x2+8x+44 by x+1.

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Chapter 15.1, Problem 6E

Chapter 15.1, Problem 8E

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Chapter 15 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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To divide: The polynomial 2 x 5 − x 2 + 8 x + 44 by x + 1 .

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Chapter 15 Solutions