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

Question

Quizplus · Accepted Answer

We can first factor out the common factor of 2:
$$2\left(x^4-1ight)$$
Then we can use the difference of squares formula to factor the expression inside the parentheses:
\begin{align*}
x^4-1 &= (x^2)^2-1^2 \
&= (x^2+1)(x^2-1)
\end{align*}
Finally, we can use the difference of squares formula again to factor the second factor:
$$x^2-1 = (x+1)(x-1)$$
Putting it all together, we get:
\begin{align*}
2\left(x^4-1ight) &= 2\left[(x^2+1)(x^2-1)ight] \
&= 2\left[(x^2+1)(x+1)(x-1)ight]
\end{align*}
Therefore, the answer is (C).

Factor Completely 2x4−22 x ^ { 4 } - 22x4−2 A) 2(x2+1)(x+1)22 \left( x ^ { 2 } + 1 \right) ( x + 1 ) ^ { 2 }2(x2+1)(x+1)2

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