Example 12 Find a polynomial $P(x)$ that satisfies $P(-1)=1, P(1)=3$, $P(3)=21, P(21)=8823$.
Solution: This problem can be solved using the Lagrange interpolation formula, but the calculation is quite cumbersome, so we will use another interpolation formula instead.

There are infinitely many polynomials that satisfy the conditions. Since we only need to find one, let's find a cubic polynomial. Let
$$
P(x)=A(x+1)(x-1)(x-3)+B(x+
$$
1) $(x-1)+C(x+1)+D$,

Let $x=-1$, we get $D=P(-1)=1$. Let $x=1$, we get $P(1)=C(1+1)+1$, so $C=1$. Let $x=3$, we get $P(3)=B(3+1)(3-1)+(3+1)+1$, so $B=$ 2. Let $x=21$, we get $P(21)=A(21+1)(21-1)(21$ $-3)+2(21+1)(21-1)+(21+1)+1$, so $A=1$.
Therefore, a polynomial that satisfies the conditions is $P(x)=(x$
$$
\begin{array}{l}
\text { +1) }(x-1)(x-3)+2(x+1)(x-1)+(x+ \\
\text { 1) }+1=x^{3}-x^{2}+3 .
\end{array}
$$