Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 377: 21

Answer

a) The rational zeroes of the function are $\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{6},\pm \frac{5}{2},\pm \frac{5}{3},\pm \frac{5}{6}$. b) The possible zeroes of the function are $-5,\frac{1}{3},\frac{1}{2}$. c) The actual zeroes of the function are $-5,\frac{1}{2},\frac{1}{3}$.

Work Step by Step

(a) According to the rational zero theorem, the possible rational zeroes are the ratio of the factor of the constant term and factor of the leading coefficient. The factors of the constant term are: $5=\pm 1,\pm 5$ And the factors of the leading coefficient are: $6=\pm 1,\pm 2,\pm 3,\pm 6$ So, the possible rational zeroes can be obtained as: $\begin{align} & \text{Possible rational zero}=\frac{\text{Factor of}\left( 5 \right)}{\text{Factor of}\left( 6 \right)} \\ & =\frac{\pm 1,\pm 5}{\pm 1,\pm 2,\pm 3,\pm 6} \\ & =\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{6},\pm \frac{5}{2},\pm \frac{5}{3},\pm \frac{5}{6} \end{align}$ There are eight possible sets of zeroes that are $\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{6},\pm \frac{5}{2},\pm \frac{5}{3},\pm \frac{5}{6}$. (b) The possible rational zeroes are $\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{6},\pm \frac{5}{2},\pm \frac{5}{3},\pm \frac{5}{6}$. Now check if -5 is an actual zero using synthetic division. $\begin{matrix} -5 & 6 & 25 & -24 & 5 \\ {} & {} & -30 & 25 & 5 \\ {} & 6 & -5 & 1 & 0 \\ \end{matrix}$ Since the remainder obtained is zero, -5 is an actual zero and the quotient is $6{{x}^{2}}-5x+1$. Therefore, the actual zeroes of the function can be $-5,\frac{1}{3},\frac{1}{2}$. (c) Using the quotient and the first factor obtained from the previous part, the rest of the zeroes can be obtained by factorizing the quotient as follows: $\begin{align} & \left( 3x-1 \right)\left( 2x-1 \right)=0 \\ & x=\frac{1}{2},\frac{1}{3} \end{align}$ Therefore, the actual zeroes of the polynomial function are $-5,\frac{1}{2},\frac{1}{3}$.
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