Factoring Polynomials

Date: Wed, 16 Nov 1994 
From: Anonymous
Subject: Factoring

I am in need of an easy method to understanding factoring polynomials,
especially with exponents.  If you can, touch a little on quadratic 
equations also. Thanks.  

Mark Muscovitch     markm@epix.net

Date: Wed, 16 Nov 1994 16:46:15 -0500 (EST)
From: Dr. Ken
Subject: Re: Factoring

Hello there!

Ah, factoring polynomials.  In my recollection, I spent pretty much my
entire 8th grade getting good at factoring polynomials.  My friend Ethan
(another Math Doctor here) spent three or four years on it, so he's really
good!

My first instinct is to tell you that the only way to really get good at
factoring polynomials is to do it a whole bunch of times.  I know that
sounds like a dumb and cheap way out, but I think it's true.  Anyway, I'll
tell you something else.

Let's say we want to factor the polynomial (3x + 2)(4x - 3) = 12x^2 - x - 6.
I'm sure you've been taught a couple of ways to do it in school, but there's
one foolproof way that always works (if your polynomial factors at all).
You can use the Quadratic Formula.  The problem is that you sometimes get
messy numbers when you use the Quadratic Formula, but sometimes you don't.
Let's see how it applies here:
                                ___________
we know that          -b +-   \/b^2 - 4ac
               x  =  ___________________________
                             2a


So we get, in our problem, 
                                _____________
                      1  +-   \/1 - 4(12)(-6)         1 +-  17     3     -2
               x  =  ___________________________ =  ___________ = ___ or ___
                             24                         24         4      3


                                           /     3  \/     2  \
                                           |x - ___ ||x + ___ |

So we think we can write our polynomial as \     4  /\     3  / .  

But look out.  You'll notice that if you actually multiply that out, you
don't get the right answer.  What you DO get, though, is the answer divided
by 12.  See that?  So what we'll do is take what we got and multiply it by
12, and we'll get (4x - 3)(3x + 2) when the dust settles.  Now THAT's the
real answer.

Was it worth it?  Could we just have made a table of all the divisors of 12,
and all the divisors of 2, and just tried with our bare hands to get the
answer?  Sure.  As a matter of fact, it would have been quicker.  But try to
factor this one:  x^2 + x + 1.  You won't be able to do it.  It doesn't
factor.  The reason is that if you look at its graph, it'll never cross the
x-axis.  So you can't look at its roots the way we did above.

What about this one:  3x^ - 2x + 17    ?  Is it messy?  Is it nice?  

I hope this helps.  More to the point, I hope I didn't confuse you even
more.  Thanks for the question!

-Ken "Dr." Math