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 4 t 4 − 17 t 2 + 14 t − 3 = 0 . Here, f ( t ) = 4 t 4 − 17 t 2 + 14 t − 3 and the degree is 4. Move from left to right, obtain the number of sign changes. From 4 t 4 → − 17 t 2 , − 17 t 2 → 14 t and 14 t → − 3 , there are 3 sign changes. Thus, the number of positive roots is 3. Change the function f ( t ) to f ( − t ) as follows. f ( − t ) = 4 ( − t ) 4 − 17 ( − t ) 2 + 14 ( − t ) − 3 = 4 t 4 − 17 t 2 − 14 t − 3 Move from left to right, obtain the number of sign changes. From 4 t 4 → − 17 t 2 , there is 1 sign changes. Hence, the maximum possible numbers of positive and negative roots of 4 t 4 − 17 t 2 + 14 t − 3 = 0 are 2 and 0, respectively. Thus, the possible rational roots by Rational Root Theorem are, r r = integer factor of − 3 integer factor of 4 = ± 1 , ± 3 ± 1 , ± 2 , ± 4 = ± 1 , ± 3 , ± 4 , ± 1 2 , ± 1 4 , ± 3 2 , ± 3 4 Begin with 1 , perform synthetic division as follows

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Author: Washington

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Chapter 15, Problem 43RE

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Chapter 15, Problem 42RE

Chapter 15, Problem 44RE

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Chapter 15 Solutions

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The equation 4 t 4 − 17 t 2 + 14 t − 3 = 0 .

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Chapter 15 Solutions