Factor completely.
-$x ^ { 4 } - 256$
A) Prime
B) $\left( x ^ { 2 } + 16 \right) ( x + 4 ) ( x - 4 )$
C) $( x + 4 ) ^ { 2 } ( x - 4 ) ^ { 2 }$
D) $\left( x ^ { 2 } - 16 \right) ( x + 4 ) ( x - 4 )$

Question

Quizplus · Accepted Answer

This expression is a difference of squares: $x^4 - 256 = (x^2)^2 - 16^2$, which factors into $(x^2 + 16)(x^2 - 16)$. Further, $x^2 - 16$ is also a difference of squares and can be factored into $(x + 4)(x - 4)$, making the complete factorization $(x^2 + 16)(x + 4)(x - 4)$.

Factor Completely x4−256x ^ { 4 } - 256x4−256 A) Prime B)

Factor Completely $x ^ { 4 } - 256$ A) Prime
B)