Answer
$P(x)=x^{2}-2x+3$
Work Step by Step
The degree is 2, there are 2 complex zeros,
$(1+i\sqrt{2})$ and $(1-i\sqrt{2})$ are conjugates,
$[x-(1+i\sqrt{2})]$ and $[x-(1-i\sqrt{2})] $are factors,
$(x-1-i\sqrt{2})$ and $(x-1+i\sqrt{2}) $are factors.
So,
$P(x)=a(x-1-i\sqrt{2})(x-1+i\sqrt{2})$
... taking $a=1$,
... applying a difference of squares,
$P(x)=[(x-1)-i\sqrt{2}][(x-1)+i\sqrt{2}]$
$=(x-1)^{2}-(i\sqrt{2})^{2}$
... apply a square of a difference and $(i^{2}=-1)$
$=x^{2}-2x+1-2(-1)$
$=x^{2}-2x+3$
$P(x)=x^{2}-2x+3$
