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

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