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