$$\begin{aligned}
f(x)= & 3 x^{14}+4 x^{13}+2 x^{11}+x^{9}+x^{6}+x^{3} \\
& +12 x^{2}+x \equiv 0(\bmod 5) .
\end{aligned}$$

Example 4 Simplify the congruence equation
$$\begin{aligned}
f(x)= & 3 x^{14}+4 x^{13}+2 x^{11}+x^{9}+x^{6}+x^{3} \\
& +12 x^{2}+x \equiv 0(\bmod 5) .
\end{aligned}$$
From the identical congruence (3) $(p=5)$, we can obtain
$$\begin{array}{ll}
x^{14} \equiv x^{6} \equiv x^{2}(\bmod 5), & x^{13} \equiv x^{5} \equiv x(\bmod 5) \\
x^{11} \equiv x^{3}(\bmod 5), & x^{9} \equiv x^{5} \equiv x(\bmod 5) \\
x^{6} \equiv x^{2}(\bmod 5) &
\end{array}$$

Therefore, the original congruence equation is equivalent to
$$3 x^{3}+16 x^{2}+6 x \equiv 0(\bmod 5),$$

which is further equivalent to
$$2\left(3 x^{3}+16 x^{2}+6 x\right) \equiv x^{3}+2 x^{2}+2 x \equiv 0(\bmod 5)$$

By direct calculation, the solutions are $x \equiv 0,1,2(\bmod 5)$. If polynomial division is used, we get
$$\begin{aligned}
f(x)= & \left(x^{5}-x\right)\left(3 x^{9}+4 x^{8}+2 x^{6}+3 x^{5}+5 x^{4}\right. \\
& \left.+2 x^{2}+4 x+5\right)+\left(3 x^{3}+16 x^{2}+6 x\right)
\end{aligned}$$

yielding the same result.