Finding a polynomial $f$ over $GF(17)$ given that $f(3)=5, f(2)=6$ and $f(8)=1$.

Question

Suppose $f$ is a polynomial over $GF(17)$ of degree at most two. Find $f$ given that $f(3)=5, f(2)=6$ and $f(8)=1$.

I have tried trial and error but had no luck, I was wondering if there was a better method? I know $GF(17)$ is a cyclic group of order $17^2-1=288$, don't know if I can use this to help me?

Answer 1

$$ f(x) = \frac{1}{30} (x-3)(x-2) + (x-3)(x-8) - (x-2)(x-8) $$

I would explain the method I used to come up with this, but I believe it is fairly obvious...