Example 1 Determine whether the congruence equation $2 x^{3}+5 x^{2}+6 x+1 \equiv 0(\bmod 7)$ has three solutions.
Solve: Here the coefficient of the first term is 2. By making an identity transformation, we can know that the original equation has the same solutions as
$$4\left(2 x^{3}+5 x^{2}+6 x+1\right) \equiv x^{3}-x^{2}+3 x-3 \equiv 0(\bmod 7)$$

Performing polynomial division, we get
$$x^{7}-x=\left(x^{3}-x^{2}+3 x-3\right)\left(x^{3}+x^{2}-2 x-2\right) x+7 x\left(x^{2}-1\right) .$$

Therefore, the number of solutions to the original congruence equation is 3.