For my math class, I was attempting to create a curve sketching question by writing the second derivative as a factorable quadratic, and working backwards to an order-4 polynomial. Along the way, I would fill in the missing constant terms by using synthetic division on an arbitrary binomial factor, and striking upon a satisfactory polynomial by trial and error. What I was aiming for was for the original f(x) polynomial, its first derivative and its second derivative to all have rational roots. After about 6-8 hours of lackluster results (the only ones that worked by this method had triple roots), I tried the internet. It was then it became clear what an ambitious project this in fact was. I had downloaded graduate-level publications which try to tackle it. This has certainly helped me in generalizing the problem, but it appears to be something bordering on unwieldy given my time constraints.

First of all, such polynomials which are differentiable and have rational roots in their first and second derivatives are called “Nice” polynomials. The impression I am getting is that these are fairly rare and difficult to find. Ones without double or triple roots have so far been next to impossible with my method, which I thought was airtight.

Here was my plan:

1) I make a second derivative as a quadratic, which has rational factors. This gives me my points of inflection for when I obtain f(x).

2) Working backwards, the first derivative is found by the indefinite integral. The result will be a polynomial with an unknown constant term. That can be found by choosing an arbitrary binomial factor and synthetic division.

- Once that is done, you need to check to see if the whole of f ‘(x) can be factored. Of course, your arbitrary factor will always work, but you might find that the quadratic which remains will not factor further.
- If you have no luck factoring the quadratic, your polynomial isn’t “nice”. Either use a different arbitrary factor or start over altogether.

3) Working backwards once more, I found my quartic by finding f(x) through the indefinite integral of f ‘(x). I add the constant term once again by synthetic division using an arbitrary factor.

- Whether the rest of it can factor is a separate question, so I must factor the remaining cubic to see if I get 3 rational roots.
- If it fails, the polynomial is not “nice”. Either pick a different arbitrary factor or start over.

You might want to make a compromise and tell yourself that you’re willing to live with a situation that when using synthetic division, you wind up with quadratics that have real roots, but everything else works out OK. Then, your students have to use the quadratic formula, which isn’t necessarily a bad thing. It’s just that I don’t think such roots should occur too early. Getting them in f(x) is fine, but not really OK for its derivatives.