Explanation Given information: The polynomial is P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 . Concept used: Descartes' Rule of Signs: Let P be a polynomial with real coefficients then, The number of positive real zeros of P either is the number of chances of sign in the coefficients of P ( x ) or is less than that by an even number. The number of negative real zeros of P either is the number of changes of sign in the coefficients of P ( − x ) or is less than that by an even number. Theorems used: Rational Zeros Theorem: If a polynomial Q ( x ) = a n x n + a n − 1 x n + ... + a 1 x 1 + a 0 x 0 has integer coefficients (here a n ≠ 0 and a 0 ≠ 0 ), then every rational zero of Qis of the form, p q Here, pis a factor of the constant coefficient a 0 and q is a factor of the leading coefficient a n . Factor theorem: If x = a is a zero of the polynomial P ( x ) , then ( x − a ) is a factor of the polynomial P ( x ) . Calculation: The degree of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 is 5 ,so it has five zeroes. To obtain possible number of positive zeroes proceed as follows, P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 Since, the polynomial P ( x ) has four variations in sign, so by Descartes' rule of sign it has four positive zeroes. Now, to obtain possible number of negative zeros substitute − x for x in P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 . P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 P ( − x ) = − x 5 − 3 x 4 + x 3 + 11 x 2 + 12 x + 4 Since, P ( − x ) has one variation in sign, so it has one negative zeros. Now, obtain the roots of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 by rational zeros theorem. In the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 , the constant coefficient term is 4 and the leading coefficient is 1. So, possible rational zero of P(x) are obtained in the form , factor of 4 factor of 1 The factors of 4 are ± 1 , ± 2 , ± 4 and factors of 1 are ± 1 . Thus, the possible rational zeros of P ( x ) are ± 1 , ± 2 , ± 4 . Now, to check the actual zeros out of the possible zeros, use synthetic division. Test whether 1 is a zero of P ( x ) or not. 1 1 − 3 − 1 11 − 12 4 1 − 2 − 3 8 − 4 _ 1 − 2 − 3 8 − 4 0 _ The remainder is 0 ,and the quotient is x 4 − 2 x 3 − 3 x 2 + 8 x − 4 . Hence 1 is a zero or ( x − 1 ) is a factor of P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 3, Problem 63E

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All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.

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Chapter 3, Problem 62E

Chapter 3, Problem 64E

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All rational, irrational, and complex zeroes of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 and state their multiplicities.

Solution Summary: The author explains the rational, irrational, and complex zeroes of the polynomial P(x) and their multiplicities.

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To find:

All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.

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Chapter 3 Solutions

All rational, irrational, and complex zeroes of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 and state their multiplicities.

Solution Summary: The author explains the rational, irrational, and complex zeroes of the polynomial P(x) and their multiplicities.

Concept explainers

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To find: All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.

Want to see the full answer?

Chapter 3 Solutions

To find:

All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.