Answer
$f(x)=-x^3+2x^2-x+2$
Work Step by Step
If $i$ is a zero of $f$ then the complex conjugate $-i$ is also a zero.
We have three zeros ($2,~i,~-i$). We can find a three-degree polynomial.
$f(x)=a(x-2)[(x-i)[x-(-i)]$
$f(x)=a(x-2)(x^2-i^2)$
$f(x)=a(x-2)(x^2+1)$
$f(x)=a(x^3+x-2x^2-2)$
$f(x)=a(x^3-2x^2+x-2)$
$a=-1$
$f(x)=-1(x^3-2x^2+x-2)$
$f(x)=-x^3+2x^2-x+2$
