Answer
$$\color{blue}{ f(x)= x^4-8x^3+30x^2-56x+13 }$$
Work Step by Step
We are given zeros and asked to write a polynomial function.
If $ 2+\sqrt{3} $ is a zero, then
$n +2+\sqrt{3} =0$
$n +\sqrt{3} = -2 $
$n=-2-\sqrt{3} $
So our factor is $\bf{(x -2-\sqrt{3} )}$
because when $x= 2+\sqrt{3} $, $(x-2-\sqrt{3} )=0$
If $2-\sqrt{3} $ is a zero, then
$n +2-\sqrt{3} =0$
$n -\sqrt{3} = -2 $
$n= -2+\sqrt{3} $
So our factor is $\bf{(x -2+\sqrt{3} )}$
because when $x=2-\sqrt{3}$, $(x -2+\sqrt{3} )=0$
*note that $ 2+\sqrt{3} $ and $2-\sqrt{3} $ are conjugates
If $2+3i $ is a zero, then
$n+2+3i =0$
$n+3i = -2 $
$n= -2-3i $
So our factor is $\bf{(x-2-3i )}$
because when $x=2+3i$, $(x-2-3i )=0$
Since $2+3i $ is a complex number, its conjugate, $2-3i $ is also a zero
$n +2-3i =0$
$n -3i = -2 $
$n= -2+3i $
So our factor is $\bf{(x-2+3i )}$
because when $x=2-3i$, $(x-2+3i )=0$
So our function is:
$f(x)=(x -2-\sqrt{3} )(x -2+\sqrt{3} )(x-2+3i )(x-2-3i )$
$f(x)=(x^2-4x+1 )(x-2+3i )(x-2-3i )$
$f(x)=(x^2-4x+1 )(x^2-4x+4-9i^2)$
Recall that $i^2=-1$
$f(x)=(x^2-4x+1 )(x^2-4x+4-9(-1))$
$f(x)=(x^2-4x+1 )(x^2-4x+4+9)$
$f(x)=(x^2-4x+1 )(x^2-4x+13)$
$$\color{blue}{ f(x)= x^4-8x^3+30x^2-56x+13 }$$
