Chapter P, Prerequisites - Section P.6 - Factoring - P.6 Exercises - Page 43: 109

$16x^{2}(x-3)(5x-9)$

Think of the equation as $3x^2y^2+x^32y4$, with $y=4x-12$. We then get: $3x^{2}(4x-12)^{2}+x^{3}(2)(4x-12)(4)=x^{2}(4x-12)[3(4x-12)+x(2)(4)]=4x^{2}(x-3)(12x-36+8x)=4x^{2}(x-3)(20x-36)=16x^{2}(x-3)(5x-9)$