Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises - Page 343: 15

Answer

$(y+z+9)(y+z-9)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ (y+z)^2-81 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the factored form of the expression above is \begin{array}{l}\require{cancel} [(y+z)+9][(y+z)-9] \\\\= (y+z+9)(y+z-9) .\end{array}
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