Intermediate Algebra (6th Edition)

$9(x+3)(x-3)$
Factoring the $GCF=9$, then the given expression, $9x^2-81$, is \begin{array}{l} 9(x^2-9) .\end{array}Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of 2 squares, then the factored form of the expression, $9(x^2-9)$, is \begin{array}{l} 9(x+3)(x-3) .\end{array}