Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Mixed Review Exercises - Page 362: 8


$x=\left\{ -1,0,1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ x^3-x=0 ,$ in factored form. Then equate each factor to zero (Zero Product Property). Finally, solve each equation. $\bf{\text{Solution Details:}}$ Factoring the $GCF=x$, the equation above is equivalent to \begin{array}{l}\require{cancel} x(x^2-1)=0 .\end{array} The expressions $ x^2 $ and $ 1 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^2-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} x[(x)^2-(1)^2]=0 \\\\= x[(x+1)(x-1)]=0 \\\\= x(x+1)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+1=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+1=0 \\\\ x=-1 \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, $ x=\left\{ -1,0,1 \right\} .$
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