Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 8 - Polynomials and Factoring - 8-4 Multiplying Soecial Cases - Practice and Problem-Solving Exercises - Page 508: 45

Answer

$a^{2}-36b^{2}$

Work Step by Step

$(a-6b)(a+6b)$ ...identify which terms correspond to $a $and $b $ in the rule for the product of a sum and difference ($(a+b)(a-b)=a^{2}-b^{2}$). $a=a,\ b=6b$ ...substitute for $a $and $b $ in the rule $(a-6b)(a+6b)=a^{2}-(6b)^{2}$ ...simplify. =$a^{2}-36b^{2}$
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