Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 57

Answer

$-16(y+2)(y-2)$

Work Step by Step

Factoring the negative $GCF= -16 $ results to $ -16(y^2-4) $. Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of 2 squares, then, \begin{array}{l} -16(y^2-4) \\= -16(y+2)(y-2) .\end{array}
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