It is when we least expect it that a student will make an observation about a conventional method that we virtually take for granted. In this case, it was factoring

, where the factors of 'ac' add up to 'b' by decomposition. In particular, I presented for consideration. The question posed was: "How can we factor this if the 3 is not a perfect square?" Pega Alerasool, an industrious student with a different view of this procedure, did not understand the conventional algorithm as it had been presented to her. It was clear that my perception of this method and hers were very different. So after explaining that the 3 or any other coefficient of the term didn't have to be a perfect square, she asked, "Why can't it be a perfect square?" That is, why couldn't the coefficient of always be a perfect square? Now I was curious and asked her to explain her approach. In essence, let then multiply both sides by 3 in order to make the coefficient of a perfect square. The advantage here is that when we apply decomposition, we do not have to worry about the positions of positioning of the constant terms.

Factor Given

Let Substitution

then Multiplication

and

the factors of –24 whose sum is –2

are –6 and 4

As with previous methods of factoring, the position of the variable and constant terms is crucial.
Here, however, we no longer have this worry.
Since the numerical coefficient of the * x* term is the same in both brackets, it does not matter where to place the constant factors, in the case –6 and 4.
Continuing this process we have

from above

factor out the g.c.f.

Division

In general, the sequence looks like this:

Factor {where the factors of 'ac' add up

to be factorable

Let { by substitution

Then {by multiplication to make the coefficient

{ of a perfect square

Now {factors of *ac* are *a* and *c* regardless of order

{factor out the g.c.f.

{division by *a*

The only real question remaining is what to do when the coefficient of in a given

question is already a perfect square. As tempting as it is to factor as is, it doesn't work as

shown below (diagram 1). We still multiply the coefficient of by itself so that we can

get the constant factors to work properly, as in (diagram 2).

4x^2 –5x +1

does not work as there are no values a and b for

the form (2x-a)(2x-b) works

Let substitution

multiplication by 4

and and

factor out the g.c.f.

division

The above was Pega's take on how factoring by decomposition. This is how it made sense to her. If it appeals to other students like Pega, then it has become an alternative method to factoring. Try it, you'll like it.