Answer
$4(x^{2}+1)(x+1)(x-1)$
Work Step by Step
Upon inspection, we find that $4$ is common to all the terms in the expression. Therefore we factor $4$ from each term of the expression,
$4x^{4}-4$
=$4(x^{4}-1)$
Next, we further factor the quadratic equation within the parenthesis:
$4(x^{4}-1)$
=$4[(x^{2})^{2}-(1)^{2}]$
=$4(x^{2}+1)(x^{2}-1)$
=$4(x^{2}+1)((x)^{2}-(1)^{2})$
=$4(x^{2}+1)[(x+1)(x-1)]$
=$4(x^{2}+1)(x+1)(x-1)$
