I recently wrote up a piece about factoring a given quartic polynomial as the composition of quadratics on my blogger page. I’ve been playing with it some more, and can now say things slightly better. On that page, though, I noted that we can assume that the quartic we start with is monic, with constant term 0, so I continue to assume that. I won’t go through much of the ‘proof’ of what follows, at it is covered on that blogger page.

**Prop:** A quartic polynomial can be factored as the composition of two quadratics iff . Moreover, when this relation on the coefficients is satisfied, there are a two-parameter family of solutions for a factorization.

Suppose the relation is satisfied, and define

The notation comes from assuming , . In the above, we have let and be free parameters. One could, alternatively, use and as the free parameters, obtaining polynomials

Clearly this requires . It’s easy enough to check that the two pairs of polynomials compose to give the original quartic.

It seems like there should be something to say about the fact that the free parameter is more restricted (positive) than if we had chosed as the free parameter (it just has to be non-zero). Sadly, I don’t know what should be said.

I’ll probably continue playing with this a little. I like thinking about quadratics coming from vertex form, , as opposed to the standard form, used above. It’s easier to see where the quadratic is, for me, in vertex form. So I’d like to get a better visual understanding of the factorization, if I can. I smell simplifying assumption and Maple animations.

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Tags: polynomial, quadratic, quartic

This entry was posted on November 16, 2008 at 9:46 pm and is filed under Play. You can follow any responses to this entry through the RSS 2.0 feed.
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November 17, 2008 at 10:18 pm |

Guess I deleted the pingback that I should have left… This post has a continuation (‘tomorrow’) at https://sumidiot.wordpress.com/2008/11/17/qq-vertex-form/