# Is there a name for this technique involving breaking a term into multiple terms?

I recently saw a solution to the quadratic equation $$x^2-5x-6=0$$ that involved re-writing the middle term, $$-5x$$, into two terms, $$x-6x$$, so that the expression could be factored and $$x$$ solved for, vis-a-vie:

$$x^2+x-6x-6=0\\x(x+1)-6(x+1)=0\\(x+1)(x-6)=0\\x=-1\quad\text{or}\quad x=6$$

Is there a name for this technique, like how “completing the square” names another technique?

• I believe this is not a general technique for solving such problems, it is merely a way of demonstrating a known answer. Commented Jun 26, 2022 at 0:49
• It's essentially a version of "factoring by [clever] grouping", although textbook examples of this technique tend to keep individual terms intact. An appropriate name here might be "factoring by [clever] re-grouping".
– Blue
Commented Jun 26, 2022 at 0:54
• Wikipedia calls this by inspection. Commented Jun 26, 2022 at 0:58
• Think I've seen it called "split" or "distribute" as in "take $x^2+x-2$ and split/distribute the constant term between the other two $= (x^2-1)+(x-1)$" but I don't have a reference handy, and I wouldn't consider it standard language, anyway.
– dxiv
Commented Jun 26, 2022 at 4:20
• The first step is called splitting the linear term; the second step is called factoring by grouping. As Selrach Dunbar's answer states, the whole procedure is called the a-c method. Commented Jun 26, 2022 at 9:42

This technique is commonly referred to as "the a-c method". I did a quick Google search for examples and came across this.

Bottom line: If a quadratic can be factored into two linear factors with integer coefficients then the A-C Method will always produce those linear factors.

The first step is not just by inspection. Rather, when using this procedure one multiplies $$a\cdot c$$ (the coefficient of $$x^2$$ by the constant) then brainstorms to find factors of this product that sum to $$b$$ (the coefficient of $$x$$).

Here $$a \cdot c = 1 \cdot (-6) = -6$$. This product has factors of $$1$$ and $$-6$$ which sum to $$b = -5$$. So one "breaks" $$-5x$$ into $$1x + (-6)x$$ and then continues as you have indicated.

Can you see how this procedure would work to factor $$x^2+x-6$$ into the product of two linear factors?

@peterwhy and @blue

Thanks, both of you.

I’ve synthesised your replies into “factoring by clever inspection”

• I once had a teacher who used to say “by intense observation “ Commented Jun 26, 2022 at 3:44
• "By prolonged observation" in the words of my differential equations prof. Although that term could refer specifically to a family of related techniques in the differential equations. Commented Jun 26, 2022 at 4:31