Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - Mid-Chapter Review - Mixed Review - Page 335: 7

Answer

$(8n+3)(8n-3)$

Work Step by Step

The expressions $ 64n^2 $ and $ 9 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 64n^2-9 ,$ 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} (8n)^2-(3)^2 \\\\= (8n+3)(8n-3) .\end{array}
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