Answer
$Q(x)=(x-1)(x+1)(x-i)(x+i)$
Zeros: $1$ , $-1$ , $i$ , $-i$,
each with multiplicity 1.
Work Step by Step
$Q(x)=(x^{2})^{2}-1^{2}= \qquad $...difference of squares...
$=(x^{2}-1)(x^{2}+1)$
$x^{2}-1= \quad $...difference of squares...$=(x-1)(x+1)$
$x^{2}+1=x^{2}-(-1)=x^{2}-i^{2} \quad $...difference of squares...$=(x-i)(x+i)$
$Q(x)=(x-1)(x+1)(x-i)(x+i)$
Zeros: $1$ , $-1$ , $i$ , $-i$,
each with multiplicity 1.