Suppose $P(x)=ax^2+bx+c$ and $P(1)=1,P(2)=2$, and $P(3)=4$. Find $a - b - c$.

Substituting the info gives us a system with 3 equations: 

\(1 = a + b + c\)

\(2 = 4a + 2b + c\)

\(3 = 16a + 4b + c\)

Subtracting the first equation from the second and the second from the third gives us another system: 

\(3a + b = 1\)

\(12a + 2b = 1\)

Solving this system gives us \(a = -{1 \over 6}\) and \(b = 1 {1 \over 2}\)

Substituting this into the first equation gives us \(c = -{1 \over 3}\), so \(a - b - c = -{1 \over 6} - {3 \over 2} + {1 \over 3} = \color{brown}\boxed{-1 {1 \over 3}}\)

That seems to be wrong....

We translate the given information about P into information about a, b and c.

\begin{align*}
P(1) &= 1 \implies a+b+c =1,\\
P(2) &= 2 \implies 4a+2b+c =2,\\
P(3) &= 4 \implies 9a+3b+c =4.
\end{align*}

Subtracting the first equation from the second and the second equation from the third gives

\begin{align*}3a+b&=1,\\
5a+b&=2.
\end{align*}

Subtracting the first equation above from the second gives 2a=1, so a=1/2 and b=-1/2. It then follows that c=1. So, our answer is\

\[\frac{1}{2} - \left(-\frac 12\right) - 1 = \boxed{0}.\]