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: 15

Answer

$\left(\dfrac{1}{3}-2z\right)\left(\dfrac{1}{3}+2z\right)$

Work Step by Step

Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then, \begin{align*} \dfrac{1}{9}-4z^2 \Rightarrow \left(\dfrac{1}{3}-2z\right)\left(\dfrac{1}{3}+2z\right) .\end{align*}
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