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: 109

Answer

$\left(\dfrac{5}{3}x^2+3y\right)\left(\dfrac{5}{3}x^2-3y\right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ \dfrac{25}{9}x^4-9y^2 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ \dfrac{25}{9}x^4 $ and $ 9y^2 $ are both perfect squares and are separated by a minus sign. Hence, $ \dfrac{25}{9}x^4-9y^2 ,$ 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} =\left(\dfrac{5}{3}x^2\right)^2 - \left(3y\right)^2 \\\\=\left(\dfrac{5}{3}x^2+3y\right)\left(\dfrac{5}{3}x^2-3y\right) .\end{array}
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