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...
 A: If you just want to use subtraction, you could take first differences 
               2   4  -3   8 
                 2  -7  11
                  -9  18
                    27

then expand this to the left
-328 -137 -36  2   4  -3   8 
   191  101  38  2  -7  11
     -90 -63 -36  -9  18
        27  27  27  27

A: Let 
$$L_3(x)=\frac{(x-4)(x-5)(x-6)}{(3-4)(3-5)(3-6)}$$
so notice that $L_3(3)=1$ and $L_3(k)=0$ for $k=4,5$ or  $6$ and define by the same way $L_4, L_5$ and $L_6$
so we see that
$$f(x)=f(3)L_3(x)+f(4)L_4(x)+f(5)L_5(x)+f(6)L_6(x)$$
and now the calculus of $f(0)$ isn't hard.
A: Let the polynomial be $f(x)=ax^3+bx^2+cx+d$. The numbers a,b,c,d are unknown.
Now, use  f(3)=2, f(4)=4, f(5)=-3, and f(6)=8. You have 4 unknowns and 4 equations.
A: 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 $f(0)=-328$.
