Example 2 Find a polynomial $p(x)$ that satisfies $p(-1)=1, p(1)=3, p(3)=21, p(21)=8823$.
Analysis: There are infinitely many polynomials that satisfy the given conditions. Since we only need to find one, we might as well find a 3rd-degree polynomial. Using the Lagrange interpolation formula, we can directly write $p(x)$, but simplification can be quite cumbersome. Instead, we can use another method to calculate, setting $p(x)=A(x+1)(x-1)(x-3)+B(x+1)(x-1)+C(x+1)+D$, and then calculate $A$, $B$, $C$, and $D$ based on the conditions.

Solution: There are infinitely many polynomials that satisfy the given conditions. Since we only need to find one, we might as well find a 3rd-degree polynomial. Set $p(x)=A(x+1)(x-1)(x-3)+B(x+1)(x-1)+C(x+1)+D$. 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, one polynomial that satisfies the given conditions is