Show that the polynomial interpolating the following data has degree three.

$\displaystyle \begin{tabular}{|r|r|}

\hline

$x$&$f(x)$\\

\hline

-2&1\\

-1&4\\

0&11\\

1&16\\

2&13\\

3&-4\\

\hline

\end{tabular}

$

If I set up a lagrange polynomial, I get $\displaystyle -x^3-x^2+7x+11$. Is this sufficient to show it is of degree 3? The official solution is $\displaystyle \Delta^3f(x_{0}) = -6, \Delta^4f(x_{0}) = 0, \Delta^5f(x_{0}) = 0$ using Newton's forward-difference formula but I don't know what delta is supposed to be.

Thanks.