## College Algebra (10th Edition)

$x^{2}(32x-9)$
We factor by grouping to obtain: $3x^{2}(8x-3)+x^{3}* 8 =3x^{2}(8x-3)+8x*x^2 =x^{2}(3(8x-3)+8x) =x^{2}(24x-9+8x) =x^{2}(32x-9)$