Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 49


$x=\left\{ -2,0,2 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ -x^3-3x^2+4x+12=0 ,$ express it first in factored form. Then equate each factor to zero using the Zero Product Property. Finally, solve each resulting equation. $\bf{\text{Solution Details:}}$ 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 \\\\ (x+3)(-1)(x^2-4)=0 \\\\ \dfrac{(x+3)(-1)(x^2-4)}{-1}=\dfrac{0}{-1} \\\\ (x+3)(x^2-4)=0 .\end{array} The expressions $ x^2 $ and $ 4 $ are both perfect squares (the square root is exact) 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), the solutions to the equation above are \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, $ x=\left\{ -2,0,2 \right\} .$
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