Let $u$ be a root of the equation
$$
x^{3}-3 x+10=0
$$

Let $f(x)$ be a quadratic polynomial with rational coefficients, and
$$
\alpha=\frac{1}{2}\left(u^{2}+u-2\right), f(\alpha)=u .
$$

Find $f(0)$.
(2010, Five Schools Joint Examination for Independent Enrollment)
From the given, we have $u^{3}-3 u+10=0$. Then
$$
\begin{aligned}
\alpha^{2} &=\frac{1}{4}\left(u^{2}+u-2\right)^{2} \\
&=\frac{1}{4}\left(u^{4}+2 u^{3}-3 u^{2}-4 u+4\right) \\
&=\frac{1}{4}\left[u(3 u-10)+2(3 u-10)-3 u^{2}-4 u+4\right] \\
&=-4-2 u .
\end{aligned}
$$

Let $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{Q}, a \neq 0)$. Then
$$
\begin{aligned}
u &=f(\alpha)=a \alpha^{2}+b \alpha+c \\
&=a(-4-2 u)+\frac{b}{2}\left(u^{2}+u-2\right)+c \\
&\Rightarrow b u^{2}+(b-4 a-2) u-(8 a+2 b-2 c)=0 . \\
\text { Let } k &=(b-4 a-2), \\
l & =-(8 a+2 b-2 c) .
\end{aligned}
$$
$$
\begin{aligned}
\text { Then } b u^{2}+k u+l &=0(k, l \in \mathbf{Q}) . \\
\text { Hence } (b u-k)\left(b u^{2}+k u+l\right) &=
\end{aligned}
$$
$$
\begin{aligned}
&=b^{2} u^{3}+\left(b l-k^{2}\right) u-k l \\
&=b^{2}(3 u-10)+\left(b l-k^{2}\right) u-k l \\
&=\left(3 b^{2}+b l-k^{2}\right) u-\left(10 b^{2}+k l\right)=0 .
\end{aligned}
$$

Since $\pm 1, \pm 2, \pm 5, \pm 10$ are not roots of equation (1), then $u$ cannot be a rational number. Therefore,
$$
\left\{\begin{array}{l}
3 b^{2}+b l-k^{2}=0 \\
10 b^{2}+k l=0 .
\end{array}\right.
$$

Assume $b \neq 0$. Then,
$$
\begin{aligned}
k^{3} &=3 b^{2} k+b k l=3 b^{2} k-10 b^{3} \\
&\Rightarrow\left(\frac{k}{b}\right)^{3}-3\left(\frac{k}{b}\right)+10=0 .
\end{aligned}
$$

Thus, $\frac{k}{b}$ is a rational root of equation (1), which is impossible, so $b=0$.
$$
\begin{aligned}
\text { By } b u^{2}+k u+l &=0 \Rightarrow k u+l=0 \\
&\Rightarrow\left\{\begin{array}{l}
k=0, \\
l=0
\end{array} \Rightarrow \left\{\begin{array}{l}
-4 a-2=0, \\
8 a-2 c=0
\end{array}\right.\right. \\
&\Rightarrow\left\{\begin{array}{l}
a=-\frac{1}{2}, \\
c=-2 .
\end{array}\right.
\end{aligned}
$$

Therefore, $f(x)=-\frac{1}{2} x^{2}-2, f(0)=-2$.
In advanced algebra, a polynomial that cannot be expressed as the product of polynomials with rational coefficients is called an irreducible polynomial over the field of rational numbers, such as $x^{3}-3 x+10$ in equation (1), which is an irreducible polynomial over the field of rational numbers. By the degree theory in advanced algebra, we know that $x^{3}-3 x+10$ is the minimal polynomial of $u$. Hence, $u^{2}, u, 1$ are linearly independent over the field of rational numbers. Therefore, from $b u^{2}+k u+l=0$, we can deduce that $b=k=l=0$, which leads to the conclusion.