College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.4 - Page 387: 12

Answer

a. $\displaystyle \pm 1, \pm 2,\pm\frac{1}{2}$ b. sample: $ 1$ is a zero...(see "step by step") c. $-\displaystyle \frac{1}{2},1,2$

Work Step by Step

see The Rational Zero Theorem: ... If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$. ------------------------ $f(x)=2x^{3}-5x^{2}+x+2$ a. candidates for zeros, $\displaystyle \frac{p}{q}:$ $p:\qquad \pm 1, \pm 2$ $q:\qquad \pm 1, \pm 2$ $\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm\frac{1}{2}$ b. Try for $x=1:$ $\begin{array}{lllll} \underline{1}| & 2 & -5 & 1 & 2\\ & & 2 & -3 & -2\\ & -- & -- & -- & --\\ & 2 & -3 & -2 & |\underline{0} \end{array}$ $1$ is a zero, $f(x)=(x-1)(2x^{2} -3x-2)$ c. Factorize the trinomial factor $(2x^{2} -3x-2)$ (find two factors of $2(-2)=-4$ whose sum is $-3):$ $(-4$ and $+1$) $2x^{2} -3x+2=2x^{2} -4x+x-2 \quad$...factor in pairs ... $=2x(x-2)+(x-2)=(2x+1)(x-2)$ $f(x)=(x-1)(2x+1)(x-2)$ The zeros of f satisfy $f(x)=0$ $(x-1)(2x+1)(x-2)=0$ $x\displaystyle \in\{-\frac{1}{2},1,2\}$
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