Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test: 10

Answer

$(3k+11j)(3k-11j)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 9k^2-121j^2 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 9k^2 $ and $ 121j^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 9k^2-121j^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} (3k)^2-(11j)^2 \\\\= (3k+11j)(3k-11j) .\end{array}
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