Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set: 54


$y=\left\{ -5,-1,1 \right\}$

Work Step by Step

The factored form of the given expression, $ x^3+5x^2=x+5 $, is \begin{array}{l} x^3+5x^2-x-5=0 \\\\ (x^3+5x^2)-(x+5)=0 \\\\ x^2(x+5)-(x+5)=0 \\\\ (x+5)(x^2-1)=0 \\\\ (x+5)(x-1)(x+1)=0 .\end{array} Equating each factor to zero, then \begin{array}{l} x+5=0 \\ x=0-5 \\ x=-5 ,\text{ OR}\\\\ x-1=0 \\ x=0+1 \\ x=1 ,\text{ OR}\\\\ x+1=0 \\ x=0-1 \\ x=-1 \end{array} Hence, $ y=\left\{ -5,-1,1 \right\} $.
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