Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Review Exercises - Page 361: 25

Answer

$(4x+5)(4x-5)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 16x^2-25 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 16x^2 $ and $ 25 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 16x^2-25 ,$ 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} (4x)^2-(5)^2 \\\\= (4x+5)(4x-5) .\end{array}
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