## Intermediate Algebra (6th Edition)

$4(x-3)(x+3)$
Factoring the $GCF= 4$ results to $4(x^2-9)$. Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then, \begin{array}{l} 4(x^2-9) \\= 4(x-3)(x+3) .\end{array}