Given an interpolating polynomial, how do I find another polynomial that interpolates at 1 less point? The polynomial 
$$p(x) = x^5-2x^4-5x^3+15x^2+4x-12$$
interpolates x = -2,-1,1,2,3,0 with p(x) = f1, f2, f3, f4, f5, -12 respectively.
In general, how do I find another polynomial of a lower degree that interpolates at all points except the last one where x = 0 without calculating f1-f5?
 A: You'll have to solve a system of linear equations. Here's the general approach.
Say that you have a polynomial
$p(x) = \displaystyle\sum_{k=0}^{n} a_k x^k$
that interpolates $n+1$ points, that is, you know all the $a_k$ values such that $p(x_i) = p_i$ for a set of $n+1$ pairs $(x_i, p_i)$, $0 \le i \le n$.
Now you want to find another polynomial
$q(x) = \displaystyle\sum_{k=0}^{n-1} b_k x^k$
that interpolates the first $n$ points from before, that is, you want to find the $b_k$ values (in terms of the $a_k$ values) so that $q(x_i) = p_i$ for $0 \le i \le n-1$.
Since you want $q(x_i) = p(x_i)$ for $0 \le i \le n-1$, then
$\underbrace{\displaystyle\sum_{k=0}^{n-1} b_k x_i^k}_{q(x_i)} = \underbrace{\displaystyle\sum_{k=0}^{n} a_k x_i^k}_{p(x_i)} = \displaystyle\sum_{k=0}^{n-1} a_k x_i^k + a_n x_i^n$
or
$\displaystyle\sum_{k=0}^{n-1} (b_k - a_k)\,x_i^k = a_n x_i^n \quad\quad (0 \le i \le n-1)$
Write this as
$\displaystyle\sum_{k=0}^{n-1} \underbrace{x_i^k}_{M_{ik}}\,\underbrace{(b_k - a_k)}_{u_k} = \underbrace{a_n x_i^n}_{v_i}$
and think of it as a matrix equation $Mu = v$. The elements of $M$ and the elements of the column vector $v$ are known and the column vector $u$ is the unknown. Once you solve this linear system, you can get the $b_k$ values from $b_k = u_k + a_k$. Note that $M$ is an $n \times n$ matrix.
