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Factoring Trinomials by the Grouping Method

Date: 12/08/2005 at 12:31:59
From: Rosemarie
Subject: Factoring trinomials by the grouping method

I would like a proof that would explain why in general factoring 
trinomials by grouping works.  For example, if factoring 21x^2 + 43x 
- 14, we need to find 2 numbers whose sum is 43 and whose product is
-294. I understand that it works (49 and -6) but not the reason behind
it, especially why the numbers must have the product of -294.  Thank
you for your help.



Date: 12/08/2005 at 13:21:33
From: Doctor Peterson
Subject: Re: Factoring trinomials by the grouping method

Hi, Rosemarie.

Let's take your example, which I'll factor by grouping:

  (1)  21x^2 + 43x - 14

  (2)  21x^2 + 49x - 6x - 14

  (3)  7x(3x + 7) - 2(3x + 7)

  (4)  (3x + 7)(7x - 2)

Working backward from the end result, look at where the numbers in 
line (2) come from:

  3x*7x + 7*7x + 3x*-2 + 7*-2

The product ac is 3*7 * 7*-2 (the coefficients in the outer terms).

The product of the middle terms is 7*3 * 7*-2, which is a product of 
the same four numbers, grouped differently.  This will always be 
true.  So the middle two numbers have to have the same product as ac; 
and they have to add up to b (43) so that line (2) is equal to line 
(1).  That is, by choosing a pair of numbers whose product is ac and 
whose sum is b, we are ensuring that the factoring by grouping will 
work.

Let's express this generally.

Suppose we're factoring

  ax^2 + bx + c

Our goal is to factor it as, say,

  (px + q)(rx + s) = prx^2 + psx + qrx + qs

So we see that

  pr = a
  ps + qr = b
  qs = c

So we want a pair of numbers, m=ps and n=qr, whose sum is b.  Notice 
that

  ac = pr * qs = ps * qr = mn

So another requirement on the two numbers m and n is that their 
product be equal to ac.

And it turns out that this is sufficient to complete the factoring. 
Once we write

  ax^2 + bx + c = ax^2 + mx + nx + c

we can factor out the GCF px from the first two terms, and factor 
out the GCF q from the second pair of terms:

  prx^2 + psx + qrx + qs = px(rx + s) + q(rx + s) = (px + q)(rx + s)

I'm assuming here that p and q will not only be factors of a, m, n, 
and c, but GCF's; that takes a little more to prove, but is true (as 
long as there is no common factor of all three coefficients, which 
can complicate things a bit).

An explanation of a related method, with more detail, can be found 
here:

  Factoring Quadratics When a Doesn't = 1
    http://mathforum.org/library/drmath/view/62562.html 

And here is a briefer explanation:

  Factoring a Trinomial
    http://mathforum.org/library/drmath/view/63894.html 

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Factoring Expressions

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