Factor completely.
-$x^{10}+32 x^{5}+256$
A) $\left(x^{5}-16\right)^{2}$
B) Prime
C) $\left(x^{5}+16\right)^{2}$
D) $\left(x^{5}+16\right)\left(x^{5}-16\right)$

Question

Quizplus · Accepted Answer

This expression is a perfect square trinomial. It can be factored as $(x^5 + 16)^2$ because $x^{10}$ is the square of $x^5$, $256$ is the square of $16$, and $2 \cdot x^5 \cdot 16 = 32x^5$, which matches the middle term.

Factor Completely x10+32x5+256x^{10}+32 x^{5}+256x10+32x5+256 A) (x5−16)2\left(x^{5}-16\right)^{2}(x5−16)2 B) Prime C) (x5+16)2\left(x^{5}+16\right)^{2}(x5+16)2 D) (x5+16)(x5−16)\left(x^{5}+16\right)\left(x^{5}-16\right)(x5+16)(x5−16)

Factor Completely $x^{10}+32 x^{5}+256$ A) $\left(x^{5}-16\right)^{2}$ B) Prime
C) $\left(x^{5}+16\right)^{2}$ D) $\left(x^{5}+16\right)\left(x^{5}-16\right)$