Intermediate Algebra (12th Edition)

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

Chapter 5 - Mixed Review Exercises - Page 362: 1

Answer

$(4+9k)(4-9k)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 16-81k^2 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 16 $ and $ 81k^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 16-81k^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} (4)^2-(9k)^2 \\\\= (4+9k)(4-9k) .\end{array}
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