Answer
$$\color{blue}{ f(x)=x^3-3x^2+2 }$$
Work Step by Step
We are given zeros and asked to write a polynomial function.
If $1- \sqrt{3} $ is a zero, then
$n +1- \sqrt{3} =0$
$n - \sqrt{3} = -1 $
$n= -1+ \sqrt{3} $
So our factor is $\bf{(x -1+ \sqrt{3} )}$
because when $x= 1- \sqrt{3} $, $(x -1+ \sqrt{3} )=0$
If $ 1+ \sqrt{3} $ is a zero, then
$n +1+\sqrt{3} =0$
$n +\sqrt{3} = -1 $
$n= -1-\sqrt{3} $
So our factor is $\bf{(x -1-\sqrt{3} )}$
because when $x= 1+ \sqrt{3} $, $(x -1-\sqrt{3} )=0$
*note that $1- \sqrt{3} $ and $ 1+ \sqrt{3} $ are conjugates
If $ 1 $ is a zero, then
$n +1 =0$
$n= -1 $
So our factor is $\bf{(x -1 )}$ because when $x= 1 $, $(x -1 )=0$
So our function is:
$f(x)=(x -1+ \sqrt{3} ) (x -1-\sqrt{3} ) (x -1 ) $
$f(x)=(x^2-x-x\sqrt{3}-x+1+\sqrt{3}+x\sqrt{3}-\sqrt{3}-3)(x -1) $
$f(x)=(x^2-2x-2)(x-1)$
$f(x)=x^3-x^2-2x^2+2x-2x+2$
$$\color{blue}{ f(x)=x^3-3x^2+2 }$$
