Take, for example, the polynomial $15x^2 + 5xy - 12y^2$.
How do I factor trinomials of the format $ax^2 + bxy + cy^2$?
6 Answers
What you have is a homogeneous polynomial of degree $2$ in $x$ and $y$ (homogeneous means that the total degree is the same in each monomial).
Divide through by $y^2$, and set $Z=\frac{x}{y}$. (This is called "de-homogeneization"). This will give you the one-variable degree $2$ polynomials $$aZ^2 + bZ + c.$$ If you know how to factor these (which is not hard), then after factoring $$aZ^2 + bZ + c = (rZ+s)(tZ+u)$$ you can multiply through by $y^2$ by multiplying one $y$ in each factor on the right hand side, to get back to the homogeneous form ("homogeneization"): $$ax^2 + bxy + cy^2 = (rx+sy)(tx+uy).$$ So it all comes down to factoring degree $2$ polynomials.
D'you know how to factor $15x^2+5x-12$? Can you see how to go from there to what you want?
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0After reading Arturo's answer, what you asked made more sense; I was thinking of the steps I took to factor ax^2 + bx + c rather than the logic behind what I did in factoring that. :) – 2012-06-26
For degree $2$ polynomials we can use the quadratic formula. This will solve for $x$ in terms of $y$.
For general $a x^2 + b x y + c y^2,$ with integers $a,b,c,$ we calculate the discriminant $$ \Delta = b^2 - 4 a c. $$ If $b^2 - 4 a c$ is nonnegative and a perfect square, such as $0,1,4,9,16, \ldots,$ then the form can be factored, otherwise not.
If you're curious, see my answer here for comparison. Note that the factorizations won't necessarily be integer factorizations. However, the methods for factoring $ax^2+bxy+cy^2$ are basically the same as those for factoring $ax^2+bx+c$, but with one extra step at the end.
I will assume that the coefficients $a$, $b$, and $c$ are real, with $a \ne 0$, and we are factoring over the reals.
We can write our polynomial as $$\frac{1}{4a}\left(4a^2x^2+4abxy+4ac\right).$$ Complete the square. We get $$\frac{1}{4a}\left((2ax+by)^2-(b^2-4ac)y^2\right).$$ If we are trying to factor over the reals, and $b^2-4ac$ is negative, there will be no factorization. But if $b^2-4ac \ge 0$, then we can factor as $$\frac{1}{4a}\left(\left(2ax+\left(b-\sqrt{b^2-4ac}\right)y\right)\left(2ax+\left(b+\sqrt{b^2-4ac}\right)y\right)\right).$$ If we wish, we can absorb $\frac{1}{4a}$ in various ways into the "main" factors.
For your particular example, the factorization is $$\frac{1}{60}\left(30x+(5-\sqrt{745})y\right)\left(30x+(5+\sqrt{745})y\right).$$