Factor completely.
-$20 \mathrm{a}^{3} \mathrm{~b}^{3}+2 \mathrm{a}^{2} \mathrm{~b}^{2}-48 \mathrm{ab}$
A) $\left(4 a^{2} b^{2}-6\right)(5 a b+8)$
B) $(4 a b+6)\left(5 a^{2} b^{2}-8\right)$
C) $a b(4 a b-6)(5 a b+8)$
D) $a b(4 a b+6)(5 a b-8)$

Question

Quizplus · Accepted Answer

First, factor out the greatest common factor (GCF), which is $ab$, resulting in $ab(20a^2b^2 + 2ab - 48)$. Further factorization of the quadratic expression in parentheses gives $ab(4ab - 6)(5ab + 8)$.

Factor Completely 20a3 b3+2a2 b2−48ab20 \mathrm{a}^{3} \mathrm{~b}^{3}+2 \mathrm{a}^{2} \mathrm{~b}^{2}-48 \mathrm{ab}20a3 b3+2a2 b2−48ab A) (4a2b2−6)(5ab+8)\left(4 a^{2} b^{2}-6\right)(5 a b+8)(4a2b2−6)(5ab+8)

Factor Completely $20 \mathrm{a}^{3} \mathrm{~b}^{3}+2 \mathrm{a}^{2} \mathrm{~b}^{2}-48 \mathrm{ab}$ A) $\left(4 a^{2} b^{2}-6\right)(5 a b+8)$