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: 49

Answer

$9w^6-z^4$

Work Step by Step

In this problem, we use the rule that $(a-b)(a+b)= a^2-b^2$. Here, $a=3w^3$ and $b=z^2$. $(3w^3)^2$ is $9w^6$ and $(z^2 )^2$ is $z^4$. Therefore, the answer is $\boxed{9w^6-z^4}$
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