Factor completely.
-$x ^ { 4 } - 1$
A) $\left( x ^ { 2 } - 1 \right) ( x + 1 ) ( x - 1 )$
B) $\left( x ^ { 2 } + 1 \right) ( x + 1 ) ( x - 1 )$
C) Prime
D) $( x + 1 ) ^ { 2 } ( x - 1 ) ^ { 2 }$

Question

Quizplus · Accepted Answer

The expression $x^4 - 1$ can be factored as a difference of squares: $x^4 - 1 = (x^2)^2 - 1^2 = (x^2 + 1)(x^2 - 1)$. Further, $x^2 - 1$ can be factored as $(x + 1)(x - 1)$, leading to the complete factorization $(x^2 + 1)(x + 1)(x - 1)$.

Factor Completely x4−1x ^ { 4 } - 1x4−1 A) (x2−1)(x+1)(x−1)\left( x ^ { 2 } - 1 \right) ( x + 1 ) ( x - 1 )(x2−1)(x+1)(x−1)

Factor Completely $x ^ { 4 } - 1$ A) $\left( x ^ { 2 } - 1 \right) ( x + 1 ) ( x - 1 )$