jmc

Setting $x=-1,$ $x=1,$ and $x=2,$ we get $$\begin{array}{rl}1=P(-1)& =P(0)-P(1)+P(2),\\ P(1)& =P(0)+P(1)+P(2),\\ P(2)& =P(0)+2P(1)+4P(2),\end{array}$$respectively. Solving this as a system of equations in $P(0),$ $P(1),$ and $P(2),$ we get $P(0)=-1,$ $P(1)=-1,$ and $P(2)=1,$ so $$P(x)=\boxed{x^2-x-1}.$$