Interpolating Polynomials

Existence

• The quadratic method is essentially the same as the linear method; join two of the points by a line and then use $m(x-1)(x-4)$ for varying values of $m$ to hit the third. The quadratic that goes through (1,0), (4,6), and (2,-2) is \[y=2x^2-8x+6\] The cubic that goes through those points and (0,-2) is \[y=x^3-5x^2+6x-2\]
• You need a polynomial of degree $n-1$. The general procedure is by induction: assume you have $(x_1,y_1)\dots(x_n,y_n)$, and that you can produce a degree $n-2$ polynomial $p$ that fits the first $n-1$ points. Then let $q(x)=\Pi_{k=1}^{n-1}(x-x_k)$, and the polynomial \[p(x)+q(x)\frac{y_n-p(x_n)}{q(x_n)}\] is degree $n-1$ and fits all $n$ points.

Uniqueness

Suppose $p(x)$ is a polynomial of degree $n$ which is zero on $n+1$ distinct points $x_1\dots x_{n+1}$. The Factor Theorem states that $p(x)=(x-x_1)\dots(x-x_{n+1})q(x)$ for some polynomial $q$. But this has degree $\ge n+1$ unless $q$ is zero, so $p$ is zero.

Suppose $r$ and $s$ are polynomials of degree n, and $r(x_i)=s(x_i)$ for $i=1\dots n+1$, $x_i$ distinct. Then $r(x)-s(x)$ is a polynomial degree $n$ which is zero in $n+1$ places, so it is zero, so $r=s$.