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.
On
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.
On
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: $p(x)=q(x-2)$.
On
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.
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.