Intermediate Algebra (6th Edition)

$12(y+3)(y-3)$
Factoring the $GCF= 12x$ results to $12x(y^2-9)$. Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of 2 squares, then, \begin{array}{l} 12x(y^2-9) \\= 12(y+3)(y-3) .\end{array}