Answer
$f(x)=x^{3}-3x^{2}+x+1$
(sample answer with $a=1$)
Work Step by Step
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
So,
$(x-1)$ is a factor of f,
$(x-(1+\sqrt{2}))=[(x-1)-\sqrt{2}]$ is a factor of f,
$(x-(1-\sqrt{2}))=[(x-1)+\sqrt{2}]$ is a factor of f.
We assume multiplicity 1, so we have
$f(x)=a(x-1)[(x-1)-\sqrt{2}] [(x-1)+\sqrt{2}]$
We are asked for $a$ polynomial, so we choose the one for which $a=1.$
$f(x)=(x-1)[(x-1)-\sqrt{2}] [(x-1)+\sqrt{2}]\quad$... Recognize a difference of squares:
$f(x)=(x-1)[(x-1)^2-(\sqrt{2})^{2}]$
$f(x)=(x-1)[x^{2}-2x+1-2]$
$ f(x)=(x-1)(x^{2}-2x-1)\quad$... distribute,
$f(x)=x(x^{2}-2x-1)-(x^{2}-2x-1)$
$f(x)=x^{3}-2x^{2}-x-x^{2}+2x+1$
$f(x)=x^{3}-3x^{2}+x+1$