[Example 1.6.5] Find all polynomials $P(x)$ that satisfy the condition
$$
x P(x-1)=(x-2) P(x), x \in \mathbf{R} .
$$
To find the analytical expression of the polynomial $P(x)$, we hope to know what its roots are, because if there is a root $x=x_{0}$, i.e., $P\left(x_{0}\right)=0$, then $P(x)$ must have a factor $x-x_{0}$. Therefore, we can substitute some special values into the known equation (1) to explore.

First, substitute $x=0$, at this time equation (1) becomes
$$
0 \cdot P(-1)=-2 P(0),
$$

thus
$$
P(0)=0,
$$

which means $P(x)$ must have a factor $x-0=x$.
Next, substitute $x=1$, at this time equation (1) becomes
$$
P(0)=-P(1),
$$

thus
$$
P(1)=-P(0)=0,
$$

which means $P(x)$ must have a factor $x-1$.
With the above results, we can assume the polynomial
$$
P(x)=x(x-1) Q(x).
$$

where $Q(x)$ is a constant or a polynomial.
Substitute (2) into equation (1) to get
$$
\begin{array}{l}
x\left[(x-1)^{2}-(x-1)\right] Q(x-1)=(x-2) x(x-1) Q(x), \\
x(x-1)(x-2) Q(x-1)=x(x-1)(x-2) Q(x).
\end{array}
$$

From this, we can get
$$
Q(x)=Q(x-1),
$$

which means
$$
Q(0)=Q(1)=Q(2)=\cdots
$$

Thus, for all real numbers $x$, $Q(x)=a$ is a constant. From equation (2), we can obtain the analytical expression of the polynomial $P(x)$
$$
P(x)=a x(x-1)
$$

which is
$$
P(x)=a x^{2}-a x.
$$

In solving this problem, the special values we chose are actually the two roots of $P(x)$, $x=0$ and $x=1$. It is these two special values that opened up the solution.