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: 53

Answer

$y=\left\{ -4,-3,3 \right\}$

Work Step by Step

The factored form of the given expression, $ y^3+4y^2=9y+36 $, is \begin{array}{l} y^3+4y^2-9y-36=0 \\\\ (y^3+4y^2)-(9y+36)=0 \\\\ y^2(y+4)-9(y+4)=0 \\\\ (y+4)(y^2-9)=0 \\\\ (y+4)(y+3)(y-3)=0 .\end{array} Equating each factor to zero, then \begin{array}{l} y+4=0 \\ y=0-4 \\ y=-4 ,\text{ OR}\\\\ y+3=0 \\ y=0-3 \\ y=-3 ,\text{ OR}\\\\ y-3=0 \\ y=0+3 \\ y=3 \end{array} Hence, $ y=\left\{ -4,-3,3 \right\} $.
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