How to find the 4th degree polynomial with given values at $0,1,2,3,4$? Determine a fourth degree polynomial p that has $p(0), p(1), p(2), p(3), p(4)$ equal to $7, 1, 3, 1, 7$, respectively. Using my ideas, I first write out the points on the polynomial as $(0,7), (1, 1),(2, 3), (4, 7).$ I assume a four degree polynomial will have the general formula $$p(x) = ax^4+bx^3+cx^2+dx+e$$ Since at $x=0$, $p(0) = 7$, it follows that $ e=7.$ 
My problem now is to find a, b, c, and d. 
 A: Symmetry is our friend, let's exploit it (that didn't come out sounding quite right.)
There is obvious symmetry about $x=2$. Symmetry about $0$ is nicer, so we first find a polynomial $q(x)$ such that $q(0)=3$, $q(\pm 1)=1$ and $q(\pm 2)=7$. Let's look for such a polynomial of shape $$q(x)=3+ax^2(x^2-1)+bx^2(x^2-4).$$
Plug in $x=2$ to find $a$ and $x=1$ to find $b$.
Then shift $q(x)$ suitably: $p(x)=q(x-2)$.  
A: Lets solve an easier case: $q(x)$ a 2nd degree polynomial and $q(1) = 1$, $q(4) = 3$ and $q(5) = 2$. It's easy to check that the following polynomial expression has the degree we want and when evaluated at 0, 4 and 5 it gives the values we expect:
$$
Q(x) = 1\cdot\frac{(x-4)(x-5)}{(1-4)(1-5)} + 3\cdot\frac{(x-5)(x-1)}{(4-5)(4-1)} + 2\cdot\frac{(x-1)(x-4)}{(5-1)(5-4)} 
$$
Therefore it must be the polynomial we are looking for, $Q(x) = q(x)$ (though, of course, you have to expand the expression to get the coefficients). Try to understand what is the trick used and then adapt it to your 4th degree polynomial. 
A: Plug in the known values and you would get $4$ (linear) equations with $4$ variables.
This is a bit much - but is doable.
If you are advanced and know what are matrices then you can use it to find the solution.
A: Yet another method, using the fact that the inputs are consecutive integers starting with zero. It’s the method of finite differences, in which you make repeated use of the operator $\Delta$, where $(\Delta f)(n)=f(n+1)-f(n)$. The basic polynomials here are $C_0(x)=1$ (constant), $C_1(x)=x$, $C_2(x)=x(x-1)/2!$, $C_3(x)=x(x-1)(x-2)/3!$, etc. In our example, I write down the original function values on the top row, and the values of $\Delta^nf$ in the succeeding rows:
\begin{align}
7\quad 1\quad3\quad1\quad7\\
-6\quad2-2\quad6\quad\>\\
8\quad-4\quad8\quad\quad\\
-12\quad12\quad\quad\quad\quad\\
24\qquad\quad\quad\quad\quad
\end{align}
Then the desired polynomial is gotten by reading down the left column: $f(x)=7C_0-6C_1+8C_2-12C_3+24C_4$.
If you are not averse to teaching a method without a supporting explanation, this is perfectly well suited to willing elementary school students.
