Explanation Given information: The polynomial P ( x ) is, P ( x ) = 4 x 4 + 2 x 3 − 2 x 2 − 3 x − 1 Concept used: Rational Zeros Theorem: Consider the polynomial P ( x ) as, P ( x ) = a n x n + a n − 1 x n − 1 + a n − 2 x n − 2 + ⋯ a 2 x 2 + a 1 x + a 0 The polynomial P ( x ) has integer coefficients, a n ≠ 0 and a 0 ≠ 0 Then every rational zero of the polynomial P ( x ) has the form j k . Here, j and k are integers. The integer j is factor of a 0 that is the constant term in the polynomial P ( x ) and the integer k is factor of a n that is leading coefficient in the polynomial P ( x ) . Formula used: The general form of quadratic equation is given as, a x 2 + b x + c = 0 Here, a ≠ 0 in the above equation. The quadratic formula used to solve the value of x from the general form of the quadratic equation is given as, x = − b ± b 2 − 4 a c 2 a Here, a , b and c are real numbers and a ≠ 0 . Calculation: The given polynomial P ( x ) is, P ( x ) = 4 x 4 + 2 x 3 − 2 x 2 − 3 x − 1 ....... (1) From Rational Zeros Theorem, the rational zeros of the polynomial P ( x ) has the form j k . In equation (1) the leading coefficient is 4 and the constant term is 1. Factors of the constant term 1 are ± 1 . Factors of the leading coefficient are ± 1 , ± 2 and ± 4 . The values of j are ± 1 and the value of k are ± 1 , ± 2 and ± 4 . Therefore, the possible rational zeros of the polynomial are, ± 1 1 , ± 1 2 , and ± 1 4 Simplify the above fractions as, ± 1 , ± 1 2 , and ± 1 4 Now, check the possible rational zeros that are zeros of the polynomial P ( x ) . Evaluate the value of P ( x ) at the possible rational zeros. Substitute 1 for x in the polynomial P ( x ) to calculate the value of P ( 1 ) as, P ( 1 ) = 4 ⋅ 1 4 + 2 ⋅ 1 3 − 2 ⋅ 1 2 − 3 ⋅ 1 − 1 = 4 + 2 − 2 − 3 − 1 = 4 − 4 = 0 Since, remainder is 0, so 1 is a zero of the polynomial P ( x ) . Therefore, ( x − 1 ) is a factor of the polynomial P ( x ) . Now, use long division method to factorize the polynomial P ( x ) as, 4 x 3 + 6 x 2 + 4 x + 1 x − 1 4 x 4 + 2 x 3 − 2 x 2 − 3 x − 1 4 x 4 − 4 x 3 _ 6 x 3 − 2 x 2 6 x 3 − 6 x 2 _ 4 x 2 − 3 x 4 x 2 − 4 x _ x − 1 x − 1 _ 0 From long division, P ( x ) factors as, P ( x ) = ( x − 1 ) ( 4 x 3 + 6 x 2 + 4 x + 1 ) Consider the cubic expression ( 4 x 3 + 6 x 2 + 4 x + 1 ) as another polynomial Q ( x ) as, Q ( x ) = 4 x 3 + 6 x 2 + 4 x + 1 ....... (2) From Rational Zeros Theorem, the rational zeros of the polynomial Q ( x ) has the form j k . In equation (2) the leading coefficient is 4 and the constant term is 1. Factors of the constant term 1 are ± 1 . Factors of the leading coefficient 4 are ± 1 , ± 2 and ± 4 . The values of j are ± 1 and the value of k are ± 1 , ± 2 and ± 4

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 3.5, Problem 62E

To determine

To calculate:

All the zeros of the polynomial P(x).

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Chapter 3.5, Problem 61E

Chapter 3.5, Problem 63E

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