Factor completely.
-$64 x^{2}-16 x+1$
A) $(8 x-1)^{2}$
B) $(8 x+1)^{2}$
C) $(8 x-1)(8 x+1)$
D) Prime

Name: Factor completely. -$64 x^{2}-16 x+1$ A) $(8 x-1)^{2}$ B) $(8 x+1)^{2}$ C) $(8 x-1)(8 x+1)$ D) Prime
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Question

Quizplus · Accepted Answer

This is a perfect square trinomial, factoring to $(8x-1)^2$, because $64x^2$ is the square of $8x$, $1$ is the square of $1$, and $-16x$ is twice the product of $8x$ and $-1$.

Factor Completely 64x2−16x+164 x^{2}-16 x+164x2−16x+1 A) (8x−1)2(8 x-1)^{2}(8x−1)2 B) (8x+1)2(8 x+1)^{2}(8x+1)2 C) (8x−1)(8x+1)(8 x-1)(8 x+1)(8x−1)(8x+1)

Factor Completely $64 x^{2}-16 x+1$ A) $(8 x-1)^{2}$ B) $(8 x+1)^{2}$ C) $(8 x-1)(8 x+1)$