I assume fundamental theorem of algebra overkills: since $P$ has degree at most $2$, it has at most $2$ roots or it is constant zero. But it's not much different either: if $P(p), P(q), P(r) = 0$, then $P(x) = (x-p)(x-q)(x-r) Q(x)$ for some polynomial $Q$. But if $Q$ is not constant zero then $P$ has degree $3$ or more, so $Q = 0$ and thus $P = 0$.