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 as follows. Since the given polynomial is of degree 4, there are at most 4 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 10 and q is the factor of 1. The possibilities of p q are, Possibilities for p Possibilities for q : ± 1 , ± 2 , ± 5 , ± 10 ± 1 Thus, the possibilities for the rational zeros of the function f ( x ) = x 4 − 6 x 3 + 9 x 2 + 6 x − 10 are ± 1 , ± 2 , ± 5 , ± 10 . 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 1 − 6 9 6 − 10 − 1 7 − 16 10 _ 1 − 7 16 − 10 0 Thus, the value of Q ( x ) = ( x 3 − 7 x 2 + 16 x − 10 ) . 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 ) b) To determine The factor of the polynomial into linear factors.

6th Edition

ISBN: 2810000027802

Author: BITTINGER

Publisher: PEARSON

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Chapter 4, Problem 61RE

To determine

The possible rational zeros and the other zeros of the function f(x)=x4−6x3+9x2+6x−10.

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The factor of the polynomial into linear factors.

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Chapter 4, Problem 60RE

Chapter 4, Problem 62RE

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The possible rational zeros and the other zeros of the function f ( x ) = x 4 − 6 x 3 + 9 x 2 + 6 x − 10 .

Solution Summary: The author explains the rational zeros of the polynomial function f(x).