Polynomial $f(x)$ degree problem. Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.
How would I solve this problem?  It seems quite complicated...

Let $g(x)=f(x+3)$.  Then the difference table for $g$ is given by
$2\;\;\; 4\;\;  -3\;\;  8$
$\;\; 2 \;\; -7\;\;  11$
$\;\;\;-9\;\;  18$
$\;\;\;\;\;\;27,$
so $f(0)=g(-3)=2+2\binom{-3}{1}-9\binom{-3}{2}+27\binom{-3}{3}=2+2(-3)-9(6)+27(-10)=-328.$

Alternatively,  $f(x)=g(x-3)=2+2\binom{x-3}{1}-9\binom{x-3}{2}+27\binom{x-3}{3}=\frac{9}{2}x^3-\frac{117}{2}x^2+245x-328$ and $f(0)=-328$.
