a) Explanation Theorem used: “ The rational zeros theorem: Let P ( x ) = a n x n + a n − 1 x n − 1 + … + a 1 x + a 0 , where all the coefficients are integers. Consider a rational number denoted by p q , where p and q are relatively prime. If p q is a zero of P ( x ) , then p is a factor of a 0 and q is the factor of a n . Calculation: Obtain the possible rational zeros of the given function f ( x ) = 2 x 5 − 13 x 4 + 32 x 3 − 38 x 2 + 22 x − 5 as follows. Since the given polynomial is of degree 5, there are at most 5 distinct zeros. From the rational zeros theorem, p q to be a zero of P ( x ) , p should be a factor of a 0 and q is the factor of a n . That is, p should be a factor of 5 and q is the factor of 2. The possibilities of p q are, Possibilities for p Possibilities for q : ± 1 , ± 5 ± 1 , ± 2 Thus, the possibilities for the rational zeros of the function f ( x ) = 2 x 5 − 13 x 4 + 32 x 3 − 38 x 2 + 22 x − 5 are ± 1 , ± 5 , ± 1 2 , ± 5 2 . Try for the above possibilities, if f ( x ) = 0 . Here, 1 satisfies the condition f ( 1 ) = 0 , hence 1 is a zero of f ( x ) . As 1 is one of the zero, it is possible to write the polynomial as, f ( x ) = ( x − 1 ) ⋅ Q ( x ) . Synthetic division is used to find the value of Q ( x ) . 2 2 − 13 32 − 38 22 − 5 2 − 11 21 − 17 5 _ 2 − 11 21 − 17 5 0 Thus, the value of Q ( x ) = ( 2 x 4 − 11 x 3 + 21 x 2 − 17 x + 5 ) . Here, 1 satisfies the condition Q ( 1 ) = 0 , hence 1 is a zero of Q ( x ) . Then, the polynomial Q ( x ) becomes, Q ( x ) = ( x − 1 ) ⋅ R ( x ) . Use the synthetic division to find the value of R ( x ) b) To determine The factor of the polynomial into linear factors.

6th Edition

ISBN: 9780135962312

Author: BITTINGER

Publisher: PEARSON

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Chapter 4, Problem 66RE

To determine

The possible rational zeros and the other zeros of the function f(x)=2x5−13x4+32x3−38x2+22x−5.

To determine

The factor of the polynomial into linear factors.

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Chapter 4, Problem 65RE

Chapter 4, Problem 67RE

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The possible rational zeros and the other zeros of the function f ( x ) = 2 x 5 − 13 x 4 + 32 x 3 − 38 x 2 + 22 x − 5 .

Solution Summary: The author explains the rational zeros of the polynomial function f(x): p and q are relatively prime.