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Polynomial of 2 variables

by Jhenrique
Tags: polynomial, variables
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Jhenrique
#1
May15-14, 05:51 AM
P: 686
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?
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UltrafastPED
#2
May15-14, 08:46 AM
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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/Quadrat...criminant.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
#3
May15-14, 09:10 AM
P: 686
Actually, I'm asking if is possible to factorize the polynomial Q(x,y)!?

pasmith
#4
May15-14, 09:39 AM
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Polynomial of 2 variables

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

That gives you six equations in six unknowns.

There is no general solution, because you can pretty quickly eliminate [itex]s = a/p[/itex], [itex]t = c/q[/itex] and [itex]u = f/r[/itex] to end up with [tex]
cp^2 + aq^2 = bpq \\
fp^2 + ar^2 = dpr \\
fq^2 + cr^2 = eqr.[/tex] These are cylinders in [itex](p,q,r)[/itex] space whose cross-sections are conic sections in the [itex](p,q)[/itex], [itex](p,r)[/itex] and [itex](q,r)[/itex] 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
#5
May15-14, 10:00 AM
P: 686
Quote Quote by pasmith View Post
[tex]
(px + qy + r)(sx + ty + u) = psx^2 + (pt + qs)xy + qty^2 + (pu + rs)x + (qu + rt)y + ru
[/tex]

That gives you six equations in six unknowns.

There is no general solution, because you can pretty quickly eliminate [itex]s = a/p[/itex], [itex]t = c/q[/itex] and [itex]u = f/r[/itex] to end up with [tex]
cp^2 + aq^2 = bpq \\
fp^2 + ar^2 = dpr \\
fq^2 + cr^2 = eqr.[/tex] These are cylinders in [itex](p,q,r)[/itex] space whose cross-sections are conic sections in the [itex](p,q)[/itex], [itex](p,r)[/itex] and [itex](q,r)[/itex] 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??


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