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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?

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  • $\begingroup$ I believe this is not a general technique for solving such problems, it is merely a way of demonstrating a known answer. $\endgroup$ Commented Jun 26, 2022 at 0:49
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    $\begingroup$ 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". $\endgroup$
    – Blue
    Commented Jun 26, 2022 at 0:54
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    $\begingroup$ Wikipedia calls this by inspection. $\endgroup$
    – peterwhy
    Commented Jun 26, 2022 at 0:58
  • $\begingroup$ 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. $\endgroup$
    – dxiv
    Commented Jun 26, 2022 at 4:20
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    $\begingroup$ 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. $\endgroup$ Commented Jun 26, 2022 at 9:42

2 Answers 2

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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?

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@peterwhy and @blue

Thanks, both of you.

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

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    $\begingroup$ I once had a teacher who used to say “by intense observation “ $\endgroup$ Commented Jun 26, 2022 at 3:44
  • $\begingroup$ "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. $\endgroup$ Commented Jun 26, 2022 at 4:31

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