Answer
$$\color{blue}{ f(x)=x^3-7x^2+12x+20 }$$
Work Step by Step
We are given zeros and asked to write a polynomial function.
If $ -1 $ is a zero, then
$n -1 =0$
$n= 1 $
So our factor is $\bf{(x+1 )}$
because when $x= -1 $, $(x+1 )=0$
If $ 4-2i $ is a zero, then
$n +4-2i =0$
$n -2i = -4 $
$n= -4+2i $
So our factor is $\bf{(x -4+2i )}$
because when $x= 4-2i $, $(x -4+2i )=0$
Since $ 4-2i $ is a complex number, its conjugate, $ 4+2i $ is also a zero
$n+4+2i =0$
$n+2i =-4$
$n= -4-2i $
So our factor is $\bf{(x-4-2i )}$
because when $x= 4+2i $, $(x-4-2i )=0$
So our function is:
$f(x)=(x+1)(x-4+2i)(x-4-2i)$
$f(x)=(x+1)(x^-8x+16-4i^2)$
Recall that $i^2=-1$
$f(x)=(x+1)(x^-8x+16-4(-1))$
$f(x)=(x+1)(x^-8x+16+4)$
$f(x)=(x+1)(x^-8x+20)$
$$\color{blue}{ f(x)=x^3-7x^2+12x+20 }$$
