Following up yesterday’s post, I’ve spent some time with factoring quartics as the composition of two quadratics, and looking at the quadratics in vertex form to get a better graphical understanding of the problem. Also, yesterday I mentioned the apparent dis-symmetry in the freedom allowed when choosing as one of the free parameters, versus choosing . I realized that came from the assumption that the leading coefficient for the quartic was positive, and probably has no further bearing on the problem.

So, let’s get back to it, as simple as possible. Our overall goal is to write as the composition where and are quadratics. Since we now know we have enough freedom in choosing some of the coefficients, without losing any power, let’s go ahead and assume that both and are monic. I think quadratics are easier to visualize in vertex form, so write , (so the vertex for is at ).

Since we’re also allowed to freely choose the constant term for , let’s choose it to be 0. This has several wonderful implications. The constant term for is , so if this is 0, we can set to eliminate a parameter. Now that the constant term for is 0, and since we’re assuming the constant term for the composite is also 0, we see that the constant term for is also 0. Thus, , and we have written . This means the vertices for the two parabolas lie on the parabola , which is somehow visually appealing. We can picture two parabolas opening up with vertices travelling around on , and make a nice movie of corresponding quartic compositions. Perhaps I’ll see what I can rig up in Maple, or something.

With these formulas for and , some things are quickly apparent. It’s comforting to check that the relation (from yesterday) is satisfied by the coefficients of the composite (here, are the coefficients of in the composite). Next, notice that

which is an even function (symmetric about the -axis). Since is just a horizontal shift of (shifting left units), this means that is symmetric around . Unless I’ve got some implications wrong somewhere, I’m pretty sure what we’ve just shown is that a quartic can be written as the composite of two quadratics precisely when it is symmetric around a vertical axis. But I’d like to double check that statement (or let you).

Suppose you had a fixed quartic (with ) and wanted to identify the and for the quadratics that we’re talking about. It’s easy to do, with the work from yesterday. Yesterday, we said that the linear coefficient of was equal to . Thus, , so . We also found that the linear coefficient of would be . So , so , which is if . Anyway, the point is that the vertices for the quadratics can be easily picked out from the coefficients of the quartic you started with.

Finally, a nice thing about the constant terms being 0 for these two quadratics is that then , so one of the roots is 0, and the other is . So the composite, will be 0 when (so ) or when (so ). It’s also not hard to find the zeros for the first and second derivatives for . The first derivative is 0 at and , and the second derivative is 0 at . Of course, not all of these values in square roots need be positive, but whatever. The symmetry around is readily apparent.

So, I’m feeling good about factoring quartics as quadratics. I’ve got no real questions remaining, personally. If you do, please let me know. What’s next? Well, there’s no point trying to factor a quintic as a composite, because 5 is prime, so the next logical choice would be a sixth degree polynomial as a quadratic and a cubic. As fun as that sounds, I may leave it to somebody else. I’ve got another project in mind in relation to polynomials, but you’ll just have to come back and find out about that later.

Tags: polynomial, quadratic, quartic, vertex form

November 18, 2008 at 12:20 am |

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