Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapters R-5 - Cumulative Review Exercises - Page 364: 31

Answer

$x=\left\{ -4,-\dfrac{3}{2},1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (x-1)(2x+3)(x+4)=0 ,$ equate each factor to zero (Zero Product Property). Then solve each resulting equation. $\bf{\text{Solution Details:}}$ Equating each factor to zero (Zero Product Property), the solutions of the equation above are \begin{array}{l}\require{cancel} x-1=0 \\\\\text{OR}\\\\ 2x+3=0 \\\\\text{OR}\\\\ x+4=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{OR}\\\\ 2x+3=0 \\\\ 2x=-3 \\\\ x=-\dfrac{3}{2} \\\\\text{OR}\\\\ x+4=0 \\\\ x=-4 .\end{array} Hence, the solutions are $ x=\left\{ -4,-\dfrac{3}{2},1 \right\} .$
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