Answer
$$\color{blue}{f(x)=x^4-4x^3+5x^2-2x-2 }$$
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-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 -1+\sqrt{2})(x-1-\sqrt{2} ) (x-1+i ) (x-1-i ) $
$f(x)=(x -1+\sqrt{2})(x-1-\sqrt{2} )(x^2-2x+1-i^2) $
Recall that $i^2=-1$
$f(x)=(x -1+\sqrt{2})(x-1-\sqrt{2} )(x^2-2x+1-(-1)) $
$f(x)=(x^2-2x+1-2 )(x^2-2x+2) $
$f(x)=(x^2-2x-1 )(x^2-2x+2) $
$$\color{blue}{f(x)=x^4-4x^3+5x^2-2x-2 }$$
