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 . Result used: Descartes’ Rule of Signs: “(a) The number of positive zeros of a polynomial equation f ( x ) = 0 cannot exceed than the number of changes in sign in f ( x ) in going from one term to the next in f ( x ) . (b) The number of negative zeros cannot exceed the number of sign changes in f ( − x ) ”. Calculation: It is given that the polynomial equation is 2 x 3 − 3 x 2 − 11 x + 6 = 0 . Here, f ( x ) = 2 x 3 − 3 x 2 − 11 x + 6 and the degree of the polynomial is 3. Move from left to right, obtain the number of sign changes. From 2 x 3 → − 3 x 2 and − 11 x → 6 , there are 2 sign changes. Thus, the number of positive roots is 2. Change the function f ( x ) to f ( − x ) as follows. f ( − x ) = 2 ( − x ) 3 − 3 ( − x ) 2 − 11 ( − x ) + 6 = − 2 x 3 − 3 x 2 + 11 x + 6 From − 3 x 2 → 11 x , there is 1 sign changes. Thus, the number of negative roots is 1. Hence, the maximum possible numbers of positive and negative roots of 3 t 3 + 8 t 2 − 1 = 0 are 2 and 1, respectively. Thus, the possible rational roots by Rational Root Theorem are as follows. r r = integer factor of 6 integer factor of 2 = ± 1 , ± 2 , ± 3 , ± 6 ± 1 , ± 2 = ± 1 , ± 2 , ± 3 , ± 6 , ± 1 2 , ± 3 2 Begin with –1, perform synthetic division as follows

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Chapter 15, Problem 40RE

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Chapter 15, Problem 39RE

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Chapter 15 Solutions

EBK BASIC TECHNICAL MATHEMATICS

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The equation 2 x 3 − 3 x 2 − 11 x + 6 = 0 .

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Chapter 15 Solutions