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is it possible to use the factor theorem when there is more than one variable? I believe so; however, don't know how to check every case.

Example:

$x^2-y^2$

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In general, a polynomial in two variables has an infinite number of zeros. Geometrically, the zeros form a curve in the plane. –  lhf Aug 31 '12 at 4:38
    
In the sense that if we treat a polynomial $p(x,y,z)$ in more than one variable as a polynomial in $x$ with coefficients involving $y$ and $z$ it reduces to the one variable case. But the best generalisations of the factor theorem lead into algebraic geometry. So (roughly) with two variables you have dimension 2, and one equation gives you a set of zeros with 1 dimension. –  Mark Bennet Aug 31 '12 at 7:46

1 Answer 1

up vote 1 down vote accepted

If you have a quadratic form with integer coefficients $a,b,c,$ as in $$ f(x,y) = a x^2 + b x y + c y^2, $$ then $f$ can be factored over the integers if and only iff $$ \Delta = b^2 - 4 a c $$ is the square of an integer. For your example $$ x^2 - y^2$$ you have $$ a = 1, \; b = 0, \; c = -1, \; \Delta = 4 = 2^2. $$

See How to factor the quadratic polynomial $2x^2-5xy-y^2$?

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Isn’t the problem of factoring a form $f(x,y)$ (in only two variables) equivalent to the problem of factoring the polynomial $f(x,1)$? –  Lubin Aug 31 '12 at 5:24
    
@will Jagy, thanks, didn't find that previous problem when I was searching here. –  yiyi Aug 31 '12 at 5:49
    
@Lubin, yes. From what I have seen, many of the students on this site are not that confident about factoring in either case. –  Will Jagy Aug 31 '12 at 6:00

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