Answer
$x=\pm1,\pm i$
$f(x)=(x+1)(x-1)(x+i)(x-i)$
Work Step by Step
Step 1. $x^4-1=0 \Longrightarrow (x^2+1)(x+1)(x-1)=0 \Longrightarrow x=\pm1\ and\ x=\pm i$
Step 2. We have $f(x)=(x+1)(x-1)(x+i)(x-i)$
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