Answer
$f(x)=x^4-10x^3+32x^2-22x-65$
Work Step by Step
If $3-2i$ is a zero of $f$ then the complex conjugate $3+2i$ is also a zero.
$f(x)=a[x-(-1)](x-5)[x-(3-2i)][x-(3+2i)]$
$f(x)=a(x+1)(x-5)[(x-3)+2i][(x-3)-2i]$
$f(x)=a(x^2-5x+x-5)[(x-3)^2-(2i)^2]$
$f(x)=a(x^2-4x-5)(x^2-6x+9+4)$
$f(x)=a(x^2-4x-5)(x^2-6x+13)$
$f(x)=a(x^4-6x^3+13x^2-4x^3+24x^2-52x-5x^2+30x-65)$
We can choose any value for $a$. So, let's make $a=1$
$f(x)=x^4-10x^3+32x^2-22x-65$
