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 3, there are at most 3 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 6 and q is the factor of 2. The possibilities of p q are, Possibilities for p Possibilities for q = ± 1 , ± 2 , ± 3 , ± 6 ± 1 , ± 2 = ± 1 ± 1 , ± 2 ± 1 , ± 3 ± 1 , ± 6 ± 1 , ± 1 ± 2 , ± 2 ± 2 , ± 3 ± 2 , ± 6 ± 2 = ± 1 , ± 2 , ± 3 , ± 6 , ± 1 2 , ± 1 , ± 3 2 , ± 3 Thus, the possibilities for the rational zeros of the function f ( x ) = 2 x 4 − 11 x 3 + 16 x 2 − x − 6 are 1 2 , − 1 2 , 3 2 , − 3 2 , 1 , − 1 , 2 , − 2 , 3 , − 3 , 6 , − 6 . Try for the above possibilities, if f ( x ) = 0 . From the graph, − 1 2 , 1 , 2 and 3 might be zeros of the polynomial. Here, − 1 2 satisfies the condition f ( − 1 2 ) = 0 , hence − 1 2 is a zero of f ( x ) . As − 1 2 is one of the zero, it is possible to write the polynomial as, f ( x ) = ( x + 1 2 ) ⋅ Q ( x ) Synthetic division is used to find the value of Q ( x ) . − 1 2 2 − 11 16 − 1 − 6 − 1 6 − 11 6 _ 2 − 12 22 − 12 0 Thus, the value of Q ( x ) = ( 2 x 3 − 12 x 2 + 22 x − 12 ) . Then, the polynomial f ( x ) becomes, f ( x ) = ( x + 1 2 ) ⋅ ( 2 x 3 − 12 x 2 + 22 x − 12 ) b) To determine The factor the polynomial into linear factors.

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ISBN: 9780135962312

Author: BITTINGER

Publisher: PEARSON

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Chapter 4, Problem 20T

To determine

The possible rational zeros and the other zeros of the function f(x)=2x4−11x3+16x2−x−6.

To determine

The factor the polynomial into linear factors.

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Chapter 4, Problem 19T

Chapter 4, Problem 21T

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The possible rational zeros and the other zeros of the function f ( x ) = 2 x 4 − 11 x 3 + 16 x 2 − x − 6 .

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