4. Let $f(x)=x^{3}-3 x^{2}+5 x-7$. Find the polynomial $g(x)$ of the smallest degree such that

$$
f(2)=g(2), \quad f(2-\sqrt{2})=g(2-\sqrt{2}), \quad f(2+\sqrt{2})=g(2+\sqrt{2})
$$
Answer: $3 x^{2}-5 x-3$.

Solution. Consider the polynomial $h(x)=g(x)-f(x)$. It is equal to zero at points $x=2$ and $x=2 \pm \sqrt{2}$. Therefore, $h(x)$ is divisible by the polynomial

\[
\begin{gathered}
q(x)=(x-2)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))= \\
=(x-2)\left((x-2)^{2}-2\right)=(x-2)\left(x^{2}-4 x+2\right)=x^{3}-6 x^{2}+10 x-4
\end{gathered}
\]

that is, for some polynomial $s(x)$, the identity

\[
g(x)-f(x)=s(x) \cdot\left(x^{3}-6 x^{2}+10 x-4\right)
\]

or

\[
g(x)=x^{3}-3 x^{2}+5 x-7+s(x) \cdot\left(x^{3}-6 x^{2}+10 x-4\right)
\]

holds. For $s(x)=-1$, we get $g(x)=3 x^{2}-5 x-3$. For any other polynomial $s(x)$, the degree of $g(x)$ is not less than three.

Evaluation. Full solution: 14 points. If the polynomial is found but the minimality of the degree is not proven, 11 points.