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Polynomial of 2 variables

  1. May 15, 2014 #1
    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. jcsd
  3. May 15, 2014 #2


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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:

    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.
  4. May 15, 2014 #3
    Actually, I'm asking if is possible to factorize the polynomial Q(x,y)!?
  5. May 15, 2014 #4


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    (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 [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.
    Last edited: May 15, 2014
  6. May 15, 2014 #5

    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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