Answer
$Q(x)=(x+i\sqrt{5 })^{2}(x-i\sqrt{5 })^{2}$
Zeros: $\pm i\sqrt{5},$
each with multiplicity 2
Work Step by Step
Let $t=x^{2}$
$t^{2}+10t+25=$ ... a square of a sum ...
$=t^{2}+2(t)(5)+5^{2}=(t+5)^{2}.$
... back-substitute:
$=(x^{2}+5)^{2}$
$Q(x)=(x^{2}+5)^{2}$
$x^{2}+5=x^{2}-5(-1)=x^{2}-5i^{2}=x^{2}-(i\sqrt{5})^{2}$
... a difference of squares ...
$=(x+i\sqrt{5 })(x-i\sqrt{5 })$
So,
$Q(x)=[(x+i\sqrt{5 })(x-i\sqrt{5 })]^{2}$
$Q(x)=(x+i\sqrt{5 })^{2}(x-i\sqrt{5 })^{2}$
Zeros: $\pm i\sqrt{5},$
each with multiplicity 2
