Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises - Page 347: 6

Answer

$(7z+4)(7z-4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 49z^2-16 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 49z^2 $ and $ 16 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 49z^2-16 ,$ 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} (7z)^2-(4)^2 \\\\= (7z+4)(7z-4) .\end{array}
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