Explanation Theorem used: Rational Root Theorem: Any rational root r r of a polynomial equation f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 0 = 0 is an integer factor of a 0 divided by an integer factor of a n . Therefore, r r = integer factor of a 0 integer factor of a n . Definition used: Descartes’ Rule of Signs: The number of positive zero of a polynomial equation f ( x ) = 0 cannot be greater than the number of changes in sign in f ( x ) in moving from one term to the next in f ( x ) . The number of negative zero cannot be greater than the number of sign changes in f ( − x ) . Calculation: It is given that the functions are y = x 4 − 11 x 2 and y = 12 x − 4 . The equation is obtained as x 4 − 11 x 2 = 12 x − 4 x 4 − 11 x 2 − 12 x + 4 = 0 Here, f ( x ) = x 4 − 11 x 2 − 12 x + 4 . Move from left to right, obtain the number of sign changes. As there are 2 sign changes, the number of positive roots is 2. Change the function f ( x ) to f ( − x ) as follows. f ( − x ) = ( − x ) 4 − 11 ( − x ) 2 − 12 ( − x ) + 4 = x 4 − 11 x 2 + 12 x + 4 There are 2 sign changes, thus the number of negative roots is 2. Hence, the maximum possible numbers of positive and negative roots of x 4 − 11 x 2 − 12 x + 4 = 0 are 2 and 2, respectively. Thus, the possible rational roots by Rational Root Theorem are, r r = integer factor of 4 integer factor of 1 = 1 , 2 , 4 1 = ± 1 , ± 2 , ± 4 Begin with a = 2 , perform synthetic division as follows. 1 0 − 11 − 12 4 − 2 4 14 − 4 1 − 2 − 7 2 0 [ − 2 ] The remainder is 0, x + 2 is the factor. Thus, the first root, a negative is –2 and the remaining factor is x 3 − 2 x 2 − 7 x + 2 . Try with a = − 2 again, perform synthetic division as follows. 1 − 2 − 7 2 − 2 8 − 2 1 − 4 1 0 [ − 2 ] The remainder is 0, x + 2 is the factor. Thus, the second root again, a negative is –2 and the remaining factor is x 2 − 4 x + 1

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 15.3, Problem 25E

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Chapter 15.3, Problem 24E

Chapter 15.3, Problem 26E

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Chapter 15 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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