2
$\begingroup$

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:

$x^2-y^2$

  • 2
    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.2012-08-31

1 Answers 1

2

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 $2x^2-5xy-y^2$?

  • 0
    @Lubin, yes. From what I have seen, many of the students on this site are not that confident about factoring in either case.2012-08-31