Answer
$P(x)=x^{2}-2x+2$
Work Step by Step
The degree is 2, there are 2 complex zeros.
$[x-(1+i)]$ and $[x-(1-i)] $are factors.
$(x-1-i)$ and $(x-1+i) $are factors.
So,
$P(x)=a(x-1-i)(x-1+i),$
... taking $a=1$,
... applying a difference of squares,
$P(x)=[(x-1)-i)][(x-1)+i)]$
$=(x-1)^{2}-i^{2}$
... apply a square of a difference and $(i^{2}=-1)$ ...
$=x^{2}-2x+1-(-1)$
$P(x)=x^{2}-2x+2$
