Factor completely.
-$108 \mathrm{a}^{4}-147 \mathrm{~b}^{2}$
A) $3\left(6 a^{2}-7 b\right)^{2}$
B) $3\left(6 a^{2}+7 b\right)\left(6 a^{2}-7 b\right)$
C) $\left(18 a^{2}+21 b\right)\left(6 a^{2}-7 b\right)$
D) $3\left(6 a^{2}+7 b\right)^{2}$

Question

Quizplus · Accepted Answer

The expression can be factored as a difference of squares: $108a^4 = (6a^2)^2$ and $147b^2 = (7b)^2$, leading to the factorization $3(6a^2 + 7b)(6a^2 - 7b)$.

Factor Completely 108a4−147 b2108 \mathrm{a}^{4}-147 \mathrm{~b}^{2}108a4−147 b2 A) 3(6a2−7b)23\left(6 a^{2}-7 b\right)^{2}3(6a2−7b)2 B) 3(6a2+7b)(6a2−7b)3\left(6 a^{2}+7 b\right)\left(6 a^{2}-7 b\right)3(6a2+7b)(6a2−7b)

Factor Completely $108 \mathrm{a}^{4}-147 \mathrm{~b}^{2}$ A) $3\left(6 a^{2}-7 b\right)^{2}$ B) $3\left(6 a^{2}+7 b\right)\left(6 a^{2}-7 b\right)$