Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Review Exercises - Page 400: 50



Work Step by Step

Given:$(x+5)^2+10(x+5)+24$ Plug $x+5=a$, we have : $a^2+10a+24$ Factorzie the equation then it can be written as: $=(a+4)(a+6)$ Since, $a=x+5$, then Hence, $(x+9)(x+11)$
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