Factoring by Grouping

I always like to understand what I'm teaching. Friday, David Elgart showed me an elegant approach to factoring rational quadratics called "Factoring by Grouping." Below I explain the technique and why it always works.

First here's a summary of the method:

Suppose you wanted to factor the trinomial ax² + bx + c into a product of two rational binomials or prove that it cannot be done.

1. Pull out the common term, possibly negative -throw it back after the last step. So WLOG (a, b, c) are relatively prime and a is positive
2. Find the factor pairs of |ac|
3. Determine what the sign of each term in the factor pair should be
4. Decide which pair to use by requiring m+n = b (in addition to mn = ab). Or if this cannot be done, stop and declare that the trinomial is irreducible.
5. (TRICKY for 8th graders) Use #3 to break up bx into mx + nx
6. (TRICKIER) Group the original trinomial using #4 as (ax² + mx) + (nx + c)
7. (SUPER TRICKY) Extract common factors and reverse the distributive law to get the final answer (aka the banana trick)

Let's illustrate the method by example. E.g., consider the trinomial 4 x² - 12 x - 27:

1. You should always pull out any common factor from a, b, and c and make sure that a is positive

   In the case of 4 x² - 12 x - 27 there was nothing to do as (4, -12, -27) are relatively prime and 4 is positive.

   
   
2. Find the factor pairs of |ac|:

   Factors of 4 * 27 = 108:
   

   • 1*108
     
   • 2*54
     
   • 3*36
   • 4*27
   • 6*18
   • 9*12

3. Determine what the sign of each term in the factor pair should be

   In this case since c is negative, one of m, n must be negative and the other positive. Since b is negative, the larger of m, n must be negative. I.e. we have the following possibilities:
   

   • 1 * -108
     
   • 2 * -54
     
   • 3 * -36
   • 4 * -27
   • 6 * -18
   • 9 * -12

   
   

4. Find numbers m, n satisfying
   
   • mn = ac
   • m+n = b
   In particular |m|, |n| will be one of the factor pairs found in part 2. Assign signs for mn based on the signs of b and c that make this possible (leaving this part vague)

   In our case b and c are both negative so that we need the smaller factor m to be positive and the bigger factor n to be negative. So our list of possibilities from #2 is:
   

   • 1,-108
     
   • 2,-54
     
   • 3,-36
   • 4,-27
   • 6,-18
   • 9,-12

   Look at the list to see which of this satisfy m + n = -12. The answer is 6, -18
   

5. If no such m,n were found in #3, the quadratic is irreducible over the integers. Otherwise we will be able to factor by grouping as follows ... - 12 x = 6x - 18 x
6. Group the original trinomial using #4 as (ax² + mx) + (nx + c) ... 4 x² - 12 x - 27 = ( 4 x² + 6x ) + ( -10 x - 27 )
7. Extract common factors and reverse the distributive law to get the final answer (aka the banana trick ) ... ( 4 x² + 6x ) + ( -18 x - 27 ) = 2 x ( 2 x + 3 ) + - 9 ( 2 x + 3 ) ... here ( 2 x + 3 ) is the "banana" (the banana trick simply says that a*banana + b*banana = (a+b)*banana .... ( 2 x - 9 ) ( 2 x + 3 ) this is the solution.

And here's the algebraic reason that this always works:

• ax² + bx + c
• ac = mn
• b = m+n

Implies

ax² + bx + c

=

ax² + mx + nx + mn/a

=

x(ax + m) + n/a(ax + m)

=

(x + n/a )(ax + m)

=

(1/a)(ax + n)(ax + m)