Explanation Given information: The polynomial function f ( x ) = 5 x 4 − 9 x 3 − 7 x 2 − 31 x − 6 . Formula used: The solution of the quadratic equation a x 2 + b x + c = 0 is x = − b ± b 2 − 4 a c 2 a . Calculation: The degree of f ( x ) is 4 . Therefore, f ( x ) has four complex zeros. By using the Descartes Rule of signs to find the real zeros, f ( x ) = 5 x 4 − 9 x 3 − 7 x 2 − 31 x − 6 , and f ( − x ) = 5 x 4 + 9 x 3 − 7 x 2 + 31 x − 6 . The number of positive real zeros of f ( x ) are 1 , as changes in the sign of the coefficient of f ( x ) are 1 . The number of negative real zeros of f ( x ) are 3 , since changes in the sign of the coefficient of f ( − x ) are 3 . By using the Rational zeros Theorem to find the potential rational zeros of f ( x ) , The factor of 6 ; all possible values of p are ± 1 , ± 2 , ± 3 , ± 6 . The factor of 5 ; all possible value of q are ± 1 , ± 5 . The potential rational zeros are ± 1 , ± 1 5 , ± 2 , ± 2 5 , ± 3 , ± 3 5 , ± 6 , ± 6 5 . r c o e f f i c i e n t o f q u o t i e n t r e m a i n d e r 1 5 − 4 − 11 − 42 − 48 − 1 5 − 9 − 2 − 40 − 8 1 5 5 − 10 − 5 − 30 0 The above table shows that 1 5 is the real zero of f ( x ) . As f ( 1 5 ) = 0 , then 1 5 is zero. Therefore, x − 1 5 is a factor of f ( x ) . By using third row in the above table, the depressed equation is f 1 ( x ) = 5 x 3 − 10 x 2 − 5 x − 30 . By using the Descartes Rule of signs to find the real zeros, f 1 ( x ) = 5 x 3 − 10 x 2 − 5 x − 30 , and f 1 ( − x ) = − 5 x 3 − 10 x 2 + 5 x − 30 . The number of positive real zeros of f 1 ( x ) are 1 , since changes in the sign of the coefficient of f 1 ( x ) are 1

11th Edition

ISBN: 9780135189405

Author: Michael Sullivan

Publisher: PEARSON

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Chapter 13, Problem 5CR

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To calculate: The complex zeros of the function f(x)=5x4−9x3−7x2−31x−6.

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Chapter 13, Problem 4CR

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The complex zeros of the function f ( x ) = 5 x 4 − 9 x 3 − 7 x 2 − 31 x − 6 .

Solution Summary: The author calculates the complex zeros of the polynomial function f(x) by using the Descartes Rule of signs.