Factoring - Continued

A smart student showed me the following trick for quick factoring. It shortens the factoring process tremendously so you don't have to do the difficult reverse distribution that factoring by grouping requires.  We start with the first four steps as in my previous blog post. But now, we use the following trick starting at a new step 5:

5. Pretend you're factoring the monic quadratic x² + bx + ac instead of  ax² + bx + c so factor it as ( x + m)( x + n ) which I'll call the "pseudo-factors"

6. Divide m by a in the first factor, and multiply x by a in the second factor obtaining ( x + m / a )( ax + n ) which is your final answer

7. You'll need to simplify.  If you're trying to get a parabola into intercept form, just pull a out of the second factor. If you're trying to get integer answers, you'll be able to pull some term from a in the second factor and use it to get rid of the simplified denominator of the first factor.

So with the example quadratic 4 x² - 12 x - 27 in my previous blog post we obtained m = 6, n = -18  in the first four steps. Apply step (5) to get pseudo-factoring ( x + 6 )( x - 18 ).  Apply step (6) with a = 4 to get the final answer  ( x + 6/4 )( 4x - 18 ) which simplifies to ( 2x + 3 )( 2x - 9) when we pull a 2 out of the second factor and multiply it into the first.

Why does this method work?  Assume that you started with the ax² + bx + c  you've found two numbers m and n with the property that mn=ac and m+n = b. The following algebra shows that the purported factors actually work.

( x + m / a )( ax + n )  

 =  ax² + ( (m / a ) a + n ) x + (m / a ) n                       (FOIL)
 =  ax² + ( m + n ) x + mn / a                                       (simplify)
 =  ax² + b x + ac / a                        ( use m+n = b and mn=ac )  
 =  ax² + b x + c                                                             (simplify)