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Given a polynomial $P(x,y)$ I would like to know what the criteria are for factoring out linear factors.

For instance, in one variable, if $Q(a) = 0$, then one may say $Q(x) = (x-a)R(x)$. In two variables this is not true, as shown by $P(x,y) = x^2+y^2$ one has $P(0,0)=0$ but one cannot factor out anything.

When can one factor out a linear factor from a polynomial in two variables?

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    @J.D. : Very relevant.2012-02-16

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I have a reference which cites this theorem for real polynomials, but if I go through the proof perhaps this also works over other fields/euclidean domains possibly. Here it is :

Theorem. Let $f(x,y)$ be a polynomial with real coefficients and degree $d$. Let $a,b,c$ be real numbers, and let $L$ denote the set of points $(x,y)$ for which $ax + by + c = 0$. If the curve $f(x,y) = 0$ and $L$ have strictly more than $d$ distinct points in common, then there exists $k(x,y)$ with real coefficients such that $ f(x,y) = (ax+by+c)k(x,y). $ This is in Niven & Zuckerman's Introduction to the theory of numbers, so it's not focused on algebra. I'll edit this answer later if I see a reason why this would hold over some other fields, but the proof uses mainly the Taylor expansion of the polynomial (which can be done formally, without using differentiation in the "limit" sense).

Hope that helps,

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    Dear Patrick, yes the result holds for every field. However the *proof* you alllude to won't work in characteristic $p$, because of the denominators in Taylor's expansion. An amusing *caveat* is that the theorem is true but empty in many cases: yes, if a polynomial of degree 17 has at least 18 zeros on a line over a field with 2 elements, the equation of the line divides the polynomial.However no line has more than 2 points (nor less) over such a field...2012-02-16
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The Factor Theorem remains applicable, for example

$\rm\qquad x-a\ |\ f(x) \iff f(a) = 0\ \ $ for $\rm\:a = by+c \in R[y]\:$ is

$\rm\qquad x-by-c\ |\ f(x,y)\iff f(by+c,\:y) = 0$

e.g. $\rm\ x-y\ |\ f(x) - f(y)\ \:$ since $\rm\ f(y)-f(y) = 0 $

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    @Patrick How does that pertain to the above?2012-02-17