1 Given the quadratic function $f(x)=x^{2}-3 x+2$, then the number of distinct real roots of the equation $f(f(x))=0$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
1) Since
$$
\begin{aligned}
f(f(x)) & =\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2 \\
& =x^{4}-6 x^{3}+10 x^{2}-3 x
\end{aligned}
$$

Therefore,
$$
x(x-3)\left(x^{2}-3 x+1\right)=0,
$$

Thus,
$$
x_{1}=0, x_{2}=3, x_{3,4}=\frac{3 \pm \sqrt{5}}{2},
$$

Hence, the original equation has 4 distinct real roots. Therefore, the answer is D.
Note: We can also discuss the distribution of the roots of $f(x)=0$. Since when $x \leqslant \frac{3}{2}$, the function $f(x)$ is monotonically decreasing, and when $x>\frac{3}{2}$, the function $f(x)$ is monotonically increasing, and the two roots of $f(x)=0$ are 1 and 2. Therefore, when $x \leqslant \frac{3}{2}$, the function $f(x) \in\left[-\frac{1}{4},+\infty\right) \supset[1,2]$, so $f(f(x))=0$ has two distinct real roots; when $x>\frac{3}{2}$, the function
$$
f(x) \in\left(-\frac{1}{4},+\infty\right) \supset[1,2],
$$

so $f(f(x))=0$ also has two distinct real roots.
In summary, the original equation has 4 distinct real roots.