Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1-5 - Cumulative Review - Page 372: 24


3, -4

Work Step by Step

We are asked to solve the equation. Thus, we first make sure the equation is set equal to zero and factor as necessary. Next, since there is 0 on one side of the equation, we set all of the factors equal to zero to solve. Doing this, we find: $$x^2+x-12=0 \\ x\left(x-3\right)+4\left(x-3\right)=0 \\ \left(x-3\right)\left(x+4\right)=0 \\ x=3,-4$$
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