I was tutoring a Year 10 student last night (he's learning about quadratics). Unfortunately, we ran into a class of problems that I couldn't explain how to solve (beyond simply guessing and checking), and this bothered me somewhat.
Some background.
My argument was that to factorize $ax^2+bx+c$, you probably shouldn't sit there "guessing and checking." After all, who wants to have to actually think when there's a relatively simple formula available? Unless the answer suggests itself to you immediately, I argued that you should immediately go ahead and use the following theorem (which is basically the quadratic formula), since:
- it solves the problem with utter reliability, and
- you don't have to think very much, and
- it doesn't rely on the original problem being "rigged" so as to admit a simple solution.
Factorization Theorem for Univariate Quadratics With Real Coefficients
Given $a,b,c \in \mathbb{R}$ and $P \in \mathbb{R}[x]$ satisfying $P = ax^2 + bx+c$, we have:
If $\Delta(P) < 0$, then $P$ is irreducible.
If $\Delta(P) \geq 0$, then $$P = a\left(x - \frac{-b-\Delta^{1/2}}{2a}\right)\left(x-\frac{-b+\Delta^{1/2}}{2a}\right)$$
(Where $\Delta(P)$ is the discriminant, which, at this level, is best defined as $b^2-4ac$.)
Of course, this is merely the "quadratic formula theorem" in a thinly-veiled disguise. However, I prefer the above version; its easier to teach, easier to use, and more "algebraic" in nature. Anyway, to cut a long story short, I taught that:
- to factorize $ax^2+bx+c$, use the above theorem.
- to solve the equation $ax^2+bx+c=0$, first factorize the LHS, and then to use the null-factor law to extract your solutions. (Hopefully, he will eventually notice that certain steps can be omitted, and thereby "discover" the quadratic formula himself; in any event, I plan not to teach it directly.)
The conundrum.
The student's book also had some bivariate problems, like:
Exercise. Factorize $x^2+2x+1-y^2$.
These problems were always rigged so as to admit an ad-hoc solution. For example:
$$x^2+2x+1-y^2 = (x+1)^2-y^2 = (x+1-y)(x+1+y)$$
However, I wanted to teach something more systematic than all this. Searching the internet for theorems/algorithms to this effect was surprisingly fruitless; everything I found either didn't address the problem directly, or else it was written at a level I didn't understand.
Question. What theorems are available to factorize bivariate quadratics with real coefficients?
Let me be more specific. Assume that we're trying to factorize $$P = ax^2+bxy+cy^2+dx+ey+f$$
I'm interested in theorems of the form: assuming certain coefficients are either $0$ or $1$,
$P$ is/isn't irreducible iff...
$P$ can be factorized as...
By making the following substitution, $P$ can be rewritten in the following, more easily factorized form...
Since students at this level have not encountered $\mathbb{C}$, hence complex numbers must be avoided. On the other hand, I have done my best to explain "formal" polynomials, so $\mathbb{R}[x,y]$ is definitely on the table.
Given $a x^2 + b xy + c y^2 + d x + e y + f = 0$:
[Complete the square to get rid of the $xy$ term.]
$( 2a x + b y )^2 - b^2 y^2 + 4a ( c y^2 + d x + e y + f ) = 0$.
Let $z = 2a x + b y$.
Then $z^2 + (4ac-b^2) y^2 + 2d z + (4ae-2bd) y + 4af = 0$.
[Complete the squares.]
$(4ac-b^2) (z+d)^2 + ( (4ac-b^2) y + (2ae-bd) )^2 = (4ac-b^2) (4af+d^2) + (2ae-bd)^2$.
[If the right-hand expression is zero, we can immediately factorize.]
[Either way, we immediately can classify into the type of conic.]