Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 5 - Section 5.3 - Special Products - Exercise Set: 36

Answer

$9y^2-\dfrac{1}{9}$

Work Step by Step

Using $(a+b)(a-b)=a^2-b^2$ or the special product of the sum and difference of like terms, the product of the given expression, $ \left( 3y+\dfrac{1}{3} \right)\left( 3y-\dfrac{1}{3} \right) ,$ is \begin{array}{l}\require{cancel} 3y(3y)-\dfrac{1}{3}\left( \dfrac{1}{3} \right) \\\\= 9y^2-\dfrac{1}{9} .\end{array}
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