Constructing a cubic given four points 
Question:  Is there an easier way to solve this 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 $f(0)$.


My Attempt: I started off with the general cubic $ax^3+bx^2+cx+d=f(x)$ and manually plugged in each point to get the following system:$$\begin{align*} & 27a+9b+3c+d=2\\ & 64a+16b+4c+d=4\\ & 125a+25b+5c+d=-3\\ & 216a+36b+6c+d=8\end{align*}\tag1$$
Solving the system with the handy matrix gives the solutions as $a=\frac 92,b=-\frac {117}2,c=245,d=-328$. Thus, $f(0)=-328$.

Even though I (think) solved the problem correctly, this method seems a bit "bulky" especially when everything becomes a higher degree. So I'm wondering if there is a quicker way to evaluate this kind of problem.
 A: There is a conceptually simpler way: do some linear algebra and polynomial arithmetic. Instead of a  complex linear system, solve $4$ much simpler linear systems. Namely,   solve the following problems: find polynomials $p(x), q(x), r(x), s(x)$ such that:
\begin{align}
(a)\enspace&\begin{cases}p(3)=1\\p(4)=0\\p(5)=0\\p(6)=0\end{cases}
&(b)\enspace&\begin{cases}q(3)=0\\q(4)=1\\q(5)=0\\q(6)=0\end{cases}
&(c)\enspace&\begin{cases}r(3)=0\\r(4)=0\\r(5)=1\\r(6)=0\end{cases}
&(d)\enspace&\begin{cases}s(3)=0\\s(4)=0\\s(5)=0\\s(6)=1\end{cases}
\end{align}
Then the solution is
$$f(x)=2p(x)+4q(x)-3r(x)+8s(x).$$
Reminder:

Let $K$ be a field, $\alpha\in K$, $f(x)$ a polynomial in $K[x]$. Then,
  $$f(\alpha)=0\iff x-\alpha\enspace\text{divides}\enspace f(x).$$

A: You can directly find out the polynomial $f$ by considering it according as the $x$-values available: 
Let $f(x)=a_0+a_1(x-3)+a_2(x-3)(x-4)+a_3(x-3)(x-4)(x-5)$ for real constants $a_0,a_1,a_2,a_3$. Note that we don't need to take into account the value $x=6$ as this is already a cubic polynomial.
Then, $f(3)=2\Rightarrow a_0=2$
$f(4)=4\Rightarrow a_1=2$
$f(5)=-3\Rightarrow a_2=-\frac{9}{2}$
$f(6)=8\Rightarrow a_3=\frac{9}{2}$
Thus, $f(x)=2+2(x-3)-\frac{9}{2}(x-3)(x-4)+\frac{9}{2}(x-3)(x-4)(x-5)$.
I think the calculations are pretty simple this way as you have chosen $f$ to be such.
A: Linear functions (degree $1$ polynomials) $p(x)$ have constant first differences $(\Delta^1f)(x)=f(x)-f(x-1)$, quadratic functions have constant second differences $(\Delta^2f)(x)=(\Delta^1f)(x)-(\Delta^1f)(x-1)$, and so on. You can use this fact to find $f(0)$ as shown below. First, from the known values, calculate the forward differences (red) from left to right, then copy the constant third difference (blue), and finally “back-calculate” to $f(0)$ (purple).

