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Polynomial of 2 variables 
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#1
May1514, 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(xx_{1})(xx_{2}), so a polynomial of 2 variables like: Q(x,y) = ax²+bxy+cy²+dx+ey+f can be written of another form?



#2
May1514, 08:46 AM

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You have a quadratic in two variables; if you plot it on the XY 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 (uh)^2/r^2 + (vg)^2/r^2 = 1, and similar for the other cases. 


#3
May1514, 09:10 AM

P: 686

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



#4
May1514, 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 crosssections 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 crosssection so that they don't intersect), and if they do all intersect they may do so at multiple points. 


#5
May1514, 10:00 AM

P: 686

I thought in something like this: ##Q(x,y) = A(xa)(xb) + B(xc)(yd) + C(ye)(yf)## Do you have more ideias?? 


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