Factor completely.
-$36 a^{3} b^{3}+45 a^{2} b^{2}-81 a b$
A) $a b(9 a b+9)(4 a b-9)$
B) $\left(9 a^{2} b^{2}-9\right)(4 a b+9)$
C) $(9 a b+9)\left(4 a^{2} b^{2}-9\right)$
D) $a b(9 a b-9)(4 a b+9)$

Question

Quizplus · Accepted Answer

The expression can be factored by first taking out the common factor $ab$, resulting in $ab(36a^2b^2 + 45ab - 81)$. Further factoring within the parentheses gives $ab(9ab - 9)(4ab + 9)$.

Factor Completely 36a3b3+45a2b2−81ab36 a^{3} b^{3}+45 a^{2} b^{2}-81 a b36a3b3+45a2b2−81ab A) ab(9ab+9)(4ab−9)a b(9 a b+9)(4 a b-9)ab(9ab+9)(4ab−9)

Factor Completely $36 a^{3} b^{3}+45 a^{2} b^{2}-81 a b$ A) $a b(9 a b+9)(4 a b-9)$