# Factoring bivariate quadratics with real coefficients (for high school students).

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.

• It's equation of second order curve (ellipse, hyperbola, parabola). We may use well-known methods for simplification of this equation. Aug 3 '15 at 5:55
• @MichaelGaluza, where does one learn this stuff? I'm not sure what to search for, or which books to look in, or which articles to read. Note also that $P$ isn't an equation, its a polynomial. Aug 3 '15 at 5:58
• You may use strightforward method: let $P = (c_1 x + c_2 y + c_3)(d_1 x + d_2 y + d_3)$ and solve a system. Aug 3 '15 at 6:19
• Solving of equations is systematic. Even if they are non-linear. Aug 3 '15 at 7:00
• Just as a tangent, keep in mind the point of the multivariate exercises is likely to get the student to recognize special-case factoring situations (namely: perfect square binomials, differences of squares, sums/differences of cubes) in nature. Pattern recognition is important! Seeing form is important! Probably the best approach for general multivariate cases (which aren't toy problems designed to have special-case-factoring solutions) that balances systematic and accessible, is to start off with Michael's approach. Them's the brakes.
– anon
Aug 3 '15 at 7:06

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.]

• I understand half the answer. I think you replace dependence on $x$ with dependence on $z$ in such a way that the coefficient of the "mixed term" $yz$ in the resulting polynomial becomes $0$. The next step makes little sense to me. If you could elaborate on how it is obtained, that would be nice. Also, note that what I really want is a factorization of $P$, not a formula for the solutions to $P=0$. I'm sure its possible to reverse-engineer the factorization given the solutions, but it would be better to answer the question as-posed rather than answering a closely-related but distinct question. Aug 3 '15 at 8:14
• @goblin: Do you know what is completing the square? Given $ax^2 + bx + c$ one can rewrite as $a(x+?)^2+?$ to get a constant times a square plus constant with respect to $x$. In my answer I multiplied first so that there is no division involved, which means it works in generalized settings like rings. So after getting rid of the $xy$ term by grouping as I did, I just completed the square to get rid of both the $x$ and $y$ terms, each one absorbed into a square term. It leaves constant terms that I collect on the right. Aug 3 '15 at 8:27
• @goblin: Also, I did answer the question completely; it tells you exactly when it can be factorized, namely when the expression I point out is zero. That case is a degenerate conic consisting of two lines. All other cases cannot be factorized but the shape is immediately obvious from the sign of $(4ac-b^2)$. Aug 3 '15 at 8:29
• What do we call the expression $4ac-b^2$ or its negative? I suppose this isn't simply called "the discriminant"... Aug 3 '15 at 8:45
• @goblin: I also didn't know what it is called, but surprisingly en.wikipedia.org/wiki/Conic_section calls its negative the discriminant. Well clearly there are many different meanings of "discriminant"! Aug 3 '15 at 8:47