Factor completely. If the polynomial cannot be factored, write "prime."
-$64 \mathrm{x}^{2}-196$
A) $4(4 x-7)^{2}$
B) $4(4 x+7)(4 x-7)$
C) $(16 x+28)(4 x-7)$
D) $(4 x+7)(16 x-28)$

Question

Quizplus · Accepted Answer

This polynomial can be factored by recognizing it as a difference of squares: $64x^2 - 196 = (8x)^2 - (14)^2$, which factors into $4(4x + 7)(4x - 7)$.

Factor Completely 64x2−19664 \mathrm{x}^{2}-19664x2−196 A) 4(4x−7)24(4 x-7)^{2}4(4x−7)2 B) 4(4x+7)(4x−7)4(4 x+7)(4 x-7)4(4x+7)(4x−7) C) (16x+28)(4x−7)(16 x+28)(4 x-7)(16x+28)(4x−7)

Factor Completely $64 \mathrm{x}^{2}-196$ A) $4(4 x-7)^{2}$ B) $4(4 x+7)(4 x-7)$ C) $(16 x+28)(4 x-7)$