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 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 cannot be greater than the number of sign changes in f ( − x ) . Calculation: It is given that the polynomial equation is x 3 + 2 x 2 − 5 x − 6 = 0 . Here, f ( x ) = x 3 + 2 x 2 − 5 x − 6 and the degree of the polynomial is 3. Move from left to right, obtain the number of sign changes. From + 2 x 2 → − 5 x , there is only 1 sign change. Thus, the number of positive roots is 1. Change the function f ( x ) to f ( − x ) as follows. f ( − x ) = ( − x ) 3 + 2 ( − x ) 2 − 5 ( − x ) − 6 = − x 3 + 2 x 2 + 5 x − 6 From − x 3 → + 2 x 2 and + 5 x → − 6 , 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 3 + 2 x 2 − 5 x − 6 = 0 are 1 and 2, respectively. Thus, the possible rational roots by Rational Root Theorem are, r r = integer factor of 6 integer factor of 1 = 1 , 2 , 3 , 6 1 = ± 1 , ± 2 , ± 3 , ± 6

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

11th Edition

ISBN: 9780137554843

Author: Allyn Washington, Richard Evans

Publisher: PEARSON+

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Ch. 15.1 - Use the factor theorem to determine whether t − 2...Ch. 15.1 - Prob. 2PE Ch. 15.1 - Prob. 3PE Ch. 15.1 - Prob. 4PE Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - In Exercises 1–4, make the given changes in the...Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E Ch. 15.1 - Prob. 6E

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