Answer
$$\color{blue}{ f(x)=x^5-2x^4+3x^3-2x^2+2x }$$
Work Step by Step
We are given zeros and asked to write a polynomial function.
If $ 0 $ is a zero, then
our factor is $\bf{(x )}$ because when $x= 0 $, $(0 )=0$
If $ i $ is a zero, then
$n+ i =0$
$n = -i $
So our factor is $\bf{(x -i )}$ because when $x= i $, $(x -i )=0$
Since $ i $ is a complex number, its conjugate, $-i$, is also a zero
$n -i =0$
$n= i $
So our factor is $\bf{(x+i )}$
because when $x= - i $, $(x+i )=0$
If $ 1+i $ is a zero, then
$n +1+i =0$
$n +i = -1 $
$n= -1-i $
So our factor is $\bf{(x -1-i )}$
because when $x= 1+i $, $(x-1-i )=0$
Since $ 1+i $ is a complex number, its conjugate, $1-i$, is also a zero
$n +1-i =0$
$n -i = -1 $
$n= -1+i $
So our factor is $\bf{(x -1+i )}$
because when $x= 1-i $, $(x-1+i )=0$
So our function is:
$f(x)=(x)(x-i)(x+i)(x-1-i)(x-1+i)$
$f(x)=(x)(x-i)(x+i)(x^2-2x+1-i^2)$
Recall that $i^2=-1$
$f(x)=(x)(x-i)(x+i)(x^2-2x+1-(-1))$
$f(x)=(x)(x-i)(x+i)(x^2-2x+2)$
$f(x)=(x)(x^2+ix-ix-i^2)(x^2-2x+2)$
$f(x)=(x)(x^2-(-1))(x^2-2x+2)$
$f(x)=(x)(x^2+1)(x^2-2x+2)$
$f(x)=(x^3+x)(x^2-2x+2)$
$f(x)=x^5-2x^4+2x^3+x^3-2x^2+2x$
$$\color{blue}{ f(x)=x^5-2x^4+3x^3-2x^2+2x }$$
