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I am looking for "fast",pencil and paper technique for factoring a bi-variate quadratic polynomial,assume the polynomial is for the form $$ax^2 + bxy + cy^2 + gx + fy + d$$

where $a,b,c,g,f,d \in \mathbb{N_0}$.

Please explain with an example.

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Do you really want the constant term to be the same as the coefficient of $y^2$? – Gerry Myerson Oct 14 '11 at 11:29
@Gerry Myerson:Sorry typo.Fixed now :) – Quixotic Oct 14 '11 at 11:30
up vote 8 down vote accepted

Let $y=0$, and factor the resulting one-variable quadratic (if possible) as $(rx+s)(r'x+s')$. Let $x=0$, and factor as $(ty+u)(t'y+u')$. Check to see whether your two factorizations are compatible. If so, they give you the factorization of the original; if not, there isn't one.

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what do you mean by compatible?Lets take this example : $x^2 - 3 x y + 2 y^2 - 3 x + 5 y + 2$,using your answer I got $(x-2) (x-1)$ and $(y+2) (2y+1)$,now I am not getting how to continue the rest? – Quixotic Oct 14 '11 at 11:42
Write that second factorization as $(-y-2)(-2y-1)$, now the two factorizations share constant terms so they are both compatible with $(x-y-2)(x-2y-1)$. – Gerry Myerson Oct 14 '11 at 11:48
Aha +1,So,if we can't write like this then we don't have any factorization right?And how about factoring something like this $15x^2-7xy-2y^2$? – Quixotic Oct 14 '11 at 11:51
I should confess that the $xy$-term still has to be checked, even if you get compatability. E.g., $x^2+79xy+2y^2-3x+5y+2$ leads to the same compatible factorizations and thus to the candidate factorization $(x-y-2)(x-2y-1)$, but when you actually multiply this out you see that the $xy$-coefficient isn't 79, so there's no factorization, despite the compatability. – Gerry Myerson Oct 14 '11 at 11:52
We crossed comments. Look at it this way: if there is a factorization, then the $x=0$ and $y=0$ factorizations must be compatible, right? So if the one-variable factorizations aren't compatible, then there's no factorization. – Gerry Myerson Oct 14 '11 at 11:54

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