Can Polynomials in Two Variables Be Expressed in Different Forms?

[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?

 

[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:
http://mathworld.wolfram.com/QuadraticCurveDiscriminant.html

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.

 

[Jhenrique]

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

 

[pasmith]

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

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 <br /> cp^2 + aq^2 = bpq \\<br /> fp^2 + ar^2 = dpr \\<br /> 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.

 

[Jhenrique]

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

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 <br /> cp^2 + aq^2 = bpq \\<br /> fp^2 + ar^2 = dpr \\<br /> 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.

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??