## Finishing Quartics

Ok, I said I was essentially done trying to factor quartics as the composition of quadratics. The one last question I had, even if I said I didn’t really have any, was if the following statement was true:

Claim: A quartic $f=x^4+bx^3+cx^2+d$ can be factored as the composition of two quadratics iff it is symmetric around $x=-b/4$.

What I have shown so far is that if it can be factored, then it is symmetric as claimed. I’ve shown that it can be factored iff $8d=4bc-b^3$. It remains to see that symmetry implies this relation on the coefficients.

Proof: If $f$ above is symmetric about $x=-b/4$, then the function $f(x-b/4)$ is even. Plugging $x-b/4$ in to $f$, expanding everything, and gathering up terms, one sees that

$f(x-b/4)=x^4+x^2(c-\frac{b^2}{4})+x(\frac{b^3}{8}-\frac{bc}{2}+d)+(\frac{b^2c}{16}-\frac{bd}{4})$

A polynomial is even if it only has even powers of $x$. Since our $f(x-b/4)$ is even, by our symmetry assumption, the linear coefficient above must be 0. If this happens, then it is easy to rearrange the equation and see that $8d=4bc-b^3$, as desired.

So there you have it. I’d summarize this work by saying that a quartic can be written as the composition of two quadratics iff it is symmetric about a vertical line.

Advertisements

Tags: ,

### One Response to “Finishing Quartics”

1. A Trickier Curve « Sumidiot’s Blog Says:

[…] should mention why I was thinking these curves, and why I want to draw that parametric curve. When I was playing with my quadric polynomials a while ago, I wanted to draw animations, but I had two parameters […]