College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises - Page 40: 108

Answer

$\left(9y+\dfrac{1}{7}\right)\left(9y-\dfrac{1}{7}\right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 81y^2-\dfrac{1}{49} ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 81y^2 $ and $ \dfrac{1}{49} $ are both perfect squares and are separated by a minus sign. Hence, $ 81y^2-\dfrac{1}{49} ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} \\\\=(9y)^2-(\frac{1}{7})^2 \\\\= \left(9y+\dfrac{1}{7}\right)\left(9y-\dfrac{1}{7}\right) .\end{array}
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