Explanation Given information: The polynomial is P ( x ) = x 5 − 4 x 4 − 3 x 3 + 22 x 2 − 4 x − 24 . Concept used: If P ( x ) = a n x n + a n − 1 x n − 1 + ... + a 1 x + a 0 has integer coefficients (where a n ≠ 0 , a 0 ≠ 0 ) then every rational zero of P is of the form, p q Here p and q are integers and p is the factor of the constant coefficient a 0 and q is a factor of leading coefficient a n . Calculation: The given polynomial is P ( x ) = x 5 − 4 x 4 − 3 x 3 + 22 x 2 − 4 x − 24 . The leading coefficient of P ( x ) is 1 and the constant term is − 24 . Consider the factors of − 24 as p such that the values of pare ± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 8 , ± 12 , ± 24 . The factors of 1 as q such that the values of q are ± 1 . Hence, the possible rational zeros are ± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 8 , ± 12 , ± 24 . Since, the polynomial P ( x ) has three variation in sign, hence it has 1 or 3 positive real zeros. To find the number of variation in sign of the polynomial P ( − x ) , substitute x = − x in the polynomial P ( x ) P ( − x ) = ( − x ) 5 − 4 ( − x ) 4 − 3 ( − x ) 3 + 22 ( − x ) 2 − 4 x − 24 = − x 5 − 4 x 4 + 3 x 3 + 22 x 2 − 4 x − 24 Since, the polynomial P ( − x ) has 2 variation in sign, hence the polynomial P ( x ) has 0 or 2 negative real zeros. Use synthetic division to check the actual zeros of the polynomial from the obtained possible zeros. To check 1 is zero or not use synthetic division as follows, 1 1 − 4 − 3 22 − 4 − 24 1 − 3 − 6 16 12 _ 1 − 3 − 6 16 12 − 12 Since remainder is not 0 , so 1 is a not a zero of P ( x ) . To check 2 is zero or not use synthetic division as follows, 2 1 − 4 − 3 22 − 4 − 24 2 − 4 − 14 16 24 _ 1 − 2 − 7 8 12 0 Here, remainder is 0 and quotient is ( x 4 − 2 x 3 − 7 x 2 + 8 x + 12 )

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 3.4, Problem 42E

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All the rational zeros of the polynomial P(x)=x5−4x4−3x3+22x2−4x−24 and express the polynomial in factored form.

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Chapter 3.4, Problem 41E

Chapter 3.4, Problem 43E

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All the rational zeros of the polynomial P ( x ) = x 5 − 4 x 4 − 3 x 3 + 22 x 2 − 4 x − 24 and express the polynomial in factored form.

Solution Summary: The author explains the rational zeros of the polynomial P(x) and the factored form of pq.

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To find:

All the rational zeros of the polynomial P(x)=x5−4x4−3x3+22x2−4x−24 and express the polynomial in factored form.

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Chapter 3 Solutions

All the rational zeros of the polynomial P ( x ) = x 5 − 4 x 4 − 3 x 3 + 22 x 2 − 4 x − 24 and express the polynomial in factored form.

Solution Summary: The author explains the rational zeros of the polynomial P(x) and the factored form of pq.

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To find: All the rational zeros of the polynomial P(x)=x5−4x4−3x3+22x2−4x−24 and express the polynomial in factored form.

Want to see the full answer?

Chapter 3 Solutions

To find:

All the rational zeros of the polynomial P(x)=x5−4x4−3x3+22x2−4x−24 and express the polynomial in factored form.