# Polynomial of 2 variables

1. May 15, 2014

### Jhenrique

If a polynomial of 1 variable, for example: P(x) = ax²+bx+c, can be written as P(x) = a(x-x1)(x-x2), so a polynomial of 2 variables like: Q(x,y) = ax²+bxy+cy²+dx+ey+f can be written of another form?

2. May 15, 2014

### UltrafastPED

You have a quadratic in two variables; if you plot it on the X-Y plane it will be a circle, ellipse, hyperbola, parabola, or a pair of lines. You can discover which by writing it in standard form and then calculating the discriminant:

Once you know the form you can rotate the coordinate system so that the cross terms disappear; use the vanishing of the cross term coefficient as the constraint.

Then put it into "standard form" for the particular geometric figure.

For a circle it will be (u-h)^2/r^2 + (v-g)^2/r^2 = 1, and similar for the other cases.

3. May 15, 2014

### Jhenrique

Actually, I'm asking if is possible to factorize the polynomial Q(x,y)!?

4. May 15, 2014

### pasmith

$$(px + qy + r)(sx + ty + u) = psx^2 + (pt + qs)xy + qty^2 + (pu + rs)x + (qu + rt)y + ru$$

That gives you six equations in six unknowns.

There is no general solution, because you can pretty quickly eliminate $s = a/p$, $t = c/q$ and $u = f/r$ to end up with $$cp^2 + aq^2 = bpq \\ fp^2 + ar^2 = dpr \\ fq^2 + cr^2 = eqr.$$ These are cylinders in $(p,q,r)$ space whose cross-sections are conic sections in the $(p,q)$, $(p,r)$ and $(q,r)$ planes respectively. There is no reason why these should all intersect (it's pretty easy to arrange three such cylinders of circular cross-section so that they don't intersect), and if they do all intersect they may do so at multiple points.

Last edited: May 15, 2014
5. May 15, 2014

### Jhenrique

Nice!

I thought in something like this:
$Q(x,y) = A(x-a)(x-b) + B(x-c)(y-d) + C(y-e)(y-f)$

Do you have more ideias??