Example 6 Let $f(x)$ represent a quartic polynomial in $x$. If $f(1)=f(2)=f(3)=0, f(4)=6$, $f(5)=72$, then the last digit of $f(2010)$ is $\qquad$. ${ }^{3}$
(2010, International Cities Mathematics Invitational for Youth)
【Analysis】It is given in the problem that $1,2,3$ are three zeros of the function $f(x)$, so we consider starting from the zero point form of the function.
Solution From the problem, we know
$$
f(1)=f(2)=f(3)=0 \text {. }
$$

Let the one-variable quartic polynomial function be
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {, }
$$

where $a, b \in \mathbf{R}$.
Therefore, $\left\{\begin{array}{l}f(4)=6(4 a+b)=6, \\ f(5)=24(5 a+b)=72 .\end{array}\right.$
Solving these, we get $a=2, b=-7$.
Thus, $f(x)=(x-1)(x-2)(x-3)(2 x-7)$.
Hence, $f(2010)$
$$
\begin{array}{l}
=(2010-1)(2010-2)(2010-3)(4020-7) \\
\equiv 2(\bmod 10) .
\end{array}
$$

Therefore, the last digit of $f(2010)$ is 2.