Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises - Page 356: 46


$x=\left\{ -2,2,3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ x^3-3x^2-4x+12=0 ,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^3-3x^2)-(4x-12)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^2(x-3)-4(x-3)=0 .\end{array} Factoring the $GCF= (x-3) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x-3)(x^2-4)=0 .\end{array} The expressions $ x^2 $ and $ 4 $ are both perfect squares and are separated by a minus sign. Hence, $ x^2-4 ,$ 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-3)(x+2)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-3=0 \\\\\text{OR}\\\\ x+2=0 \\\\\text{OR}\\\\ x-2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-3=0 \\\\ x=3 \\\\\text{OR}\\\\ x+2=0 \\\\ x=-2 \\\\\text{OR}\\\\ x-2=0 \\\\ x=2 .\end{array} Hence, the solutions are $ x=\left\{ -2,2,3 \right\} .$
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