Answer
$Q(x)=x^{3}+x$
Work Step by Step
The conjugate of i is also a zero (Conjugate Roots Theorem)
so we have three zeros, 0, $i,$ and $-i$ for a polynomial of degree 3, as there should be.
Factorization of Q(x):
$Q(x)=a(x-0)(x+i)(x-i)$
... taking a=1, apply difference of squares...
$=x(x^{2}-(-1))=x(x^{2}+1)$
$=x^{3}+x$