Example 1 Consider the equation in $x$
$$
(1+a) x^{4}+x^{3}-(3 a+2) x^{2}-4 a=0
$$
(1) Does the equation always have a real solution for any real number $a$?
(2) Is there a real number $x_{0}$ such that, regardless of what real number $a$ is, $x_{0}$ is never a solution to this equation?
Solution: We just need to rewrite the original equation as
$$
\left(x^{4}-3 x^{2}-4\right) a+\left(x^{4}+x^{3}-2 x^{2}\right)=0
$$

The problem is then solved.
(1) From $\left\{\begin{array}{l}x^{4}-3 x^{2}-4=0 \\ x^{4}+x^{3}-2 x^{2}=0\end{array}\right.$
we get $x=-2$,
so the given equation always has a real solution $x=-2$.
(2) From $\left\{\begin{array}{l}x^{4}-3 x^{2}-4=0 \\ x^{4}+x^{3}-2 x^{2} \neq 0\end{array}\right.$
we get $x_{0}=2$.
Therefore, there exists a real number $x_{0}=2$, which is never a solution to the original equation regardless of the value of $a$.