I'm trying to come up with a general strategy for factoring expressions with four terms on the basis of the symmetries of the expressions. One thought I had was the following: count up the number of terms in which each variable appears, and compare.
For example, if I want to factor the expression $x^{2}-y^{2}-4x+4$, I could say that $x$ appears in two terms, and $y$ appears in only one term, and there is only one constant term. Since the two variables do not appear in the same number of terms, I would group by 1 and 3. So I would have
$$x^{2}-4x+4-y^{2}=(x-2)^{2}-y^{2}=(x-2-y)(x-2+y).$$
On the other hand, suppose I had to factor the expression $2y-6-3x+xy$, I could note that $x$ and $y$ each appear in exactly two terms, so I should group by 2 and 2. Indeed,
$$2y-6-3x+xy=2(y-3)+x(y-3)=(x+2)(y-3).$$
However, I find that this strategy does not always work. For example, suppose I have something like $x^{2}-2xy+y^{2}-9$. Even though $x$ and $y$ appear in the same number of terms, the correct thing to do here is
$$x^{2}-2xy+y^{2}-9=(x-y)^{2}-3^{2}=(x-y-3)(x-y+3).$$
Can my strategy be salvaged? Is there a more general method than simply trial-and-error?
Important note: I am interested in basic techniques teachable in High School Algebra I.
Best Answer
Your goal is possible (modulo your students' facility with algebra) via successive reductions. In case it's helpful to you conceptually, "secretly" you're diagonalizing a quadratic form in the plane, then completing the square, and factoring.
Let's say the original expression to be factored is $ax^2 + 2bxy + cy^2 + dx + ey + f$. (N.B. factor of $2$ in the $xy$ coefficient.) If $b = 0$, skip to the next step. If $b \neq 0$, calculate the "magic numbers" (eigenvalues of the quadratic form) $$g = (1/2)\Bigl(a + c + \sqrt{(a - c)^2 + 4b^2}\Bigr),\qquad h = (1/2)\Bigl(a + c - \sqrt{(a - c)^2 + 4b^2}\Bigr).$$ Introduce new variables $$x' = \frac{(h-a)x - by}{b(h-g)},\qquad y' = \frac{(-g+a)x + by}{b(h-g)}.$$ The expression to be factored now has the form $g(x')^2 + h(y')^2 + d'x' + e'y' + f$ for some numbers $d'$ and $e'$.
We may as well assume $gh < 0$: If $g = 0$ or $h = 0$, "inspection" succeeds; if $gh > 0$, the quadratic defines an ellipse, and does not factor over the reals. Complete the square with the $x^2$ and $x$ terms and with the $y^2$ and $y$ terms, obtaining new variables $x''$ and $y''$. The equation to be factored now has the form $g(x'')^2 + h(y'')^2 + f' = 0$ for some number $f'$. This equation factors as a difference of squares because $gh < 0$.
Naturally, I've made a good faith effort to eliminate algebra errors and typos, but it's a good idea to test a few examples before using this in class. :)