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