Answer
$$\color{blue}{f(x)= x^3-8x^2+22x-20 }$$
Work Step by Step
We are given zeros and asked to write a polynomial function.
If $ 2 $ is a zero, then
$n +2 =0$
$n= -2 $
So our factor is $\bf{(x-2 )}$
because when $x= 2 $, $(x -2 )=0$
If $ 3+i $ is a zero, then
$n +3+i =0$
$n +i = -3 $
$n= -3-i $
So our factor is $\bf{(x -3-i )}$
because when $x= 3+i $, $(x-3-i )=0$
Since $ 3+i $ is a complex number, its conjugate, $ 3-i $ is also a zero
$n +3-i =0$
$n -i =-3$
$n= -3+i $
So our factor is $\bf{(x -3+i )}$
because when $x= 3-i $, $(x -3+i )=0$
So our function is:
$f(x)=(x-2)(x -3-i)(x -3+i)$
$f(x)=(x-2)(x^2-6x+9-i^2)$
Recall that $i^2=-1$
$f(x)=(x-2)(x^2-6x+9-(-1))$
$f(x)=(x-2)(x^2-6x+10)$
$$\color{blue}{f(x)= x^3-8x^2+22x-20 }$$