Factor completely.
-$8 a^{4}-98 b^{2}$
A) $\left(4 a^{2}+14 b\right)\left(2 a^{2}-7 b\right)$
B) $2\left(2 a^{2}+7 b\right)^{2}$
C) $2\left(2 a^{2}+7 b\right)\left(2 a^{2}-7 b\right)$
D) $2\left(2 a^{2}-7 b\right)^{2}$

Question

Quizplus · Accepted Answer

This expression can be factored as a difference of squares: $8a^4 - 98b^2 = (2a^2)^2 - (7b)^2$, which factors into $(2a^2 + 7b)(2a^2 - 7b)$. Factoring out a 2 gives $2(2a^2 + 7b)(2a^2 - 7b)$.

Factor Completely 8a4−98b28 a^{4}-98 b^{2}8a4−98b2 A) (4a2+14b)(2a2−7b)\left(4 a^{2}+14 b\right)\left(2 a^{2}-7 b\right)(4a2+14b)(2a2−7b) B) 2(2a2+7b)22\left(2 a^{2}+7 b\right)^{2}2(2a2+7b)2

Factor Completely $8 a^{4}-98 b^{2}$ A) $\left(4 a^{2}+14 b\right)\left(2 a^{2}-7 b\right)$ B) $2\left(2 a^{2}+7 b\right)^{2}$