Factor completely.
-$12 a ^ { 4 } - 27 b ^ { 2 }$
A) $\left( 6 a ^ { 2 } + 9 b \right) \left( 2 a ^ { 2 } - 3 b \right)$
B) $3 \left( 2 a ^ { 2 } + 3 b \right) \left( 2 a ^ { 2 } - 3 b \right)$
C) $3 \left( 2 a ^ { 2 } + 3 b \right) ^ { 2 }$
D) $3 \left( 2 a ^ { 2 } - 3 b \right) ^ { 2 }$

Question

Quizplus · Accepted Answer

The expression $12 a ^ { 4 } - 27 b ^ { 2 }$ can be factored by first taking out the greatest common factor, which is 3, resulting in $3(4a^4 - 9b^2)$. This expression is a difference of squares, which can be factored further into $3(2a^2)^2 - (3b)^2$, leading to the final factorization of $3(2a^2 + 3b)(2a^2 - 3b)$.

Factor Completely 12a4−27b212 a ^ { 4 } - 27 b ^ { 2 }12a4−27b2

Factor Completely $12 a ^ { 4 } - 27 b ^ { 2 }$