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: 58

Answer

$-12(y+3)(y-3)$

Work Step by Step

Factoring the negative $GCF= -12 $ results to $ -12(y^2-9) $. Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of 2 squares, then, \begin{array}{l} -12(y^2-9) \\= -12(y+3)(y-3) .\end{array}
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