Factoring by Grouping to Solve a Polynomial Equation

Date: 05/26/2005 at 21:06:47
From: Kate
Subject: re: Factoring and Solving Polynomial Equations

How do you factor polynomials when there is an x cubed in the equation
and it doesn't fit into a perfect square form?

For example: x^3 + 3x^2 - x - 3 = 0

I think that you would first have to add 3 to both sides:

  x^3 + 3x^2 - x = 3

Then you would be able to take an x out of the equation:

  x(x^2 + 3x - 1) = 3

The problem is that there are no factors of -1 which will work
together to give you a positive 3 in the middle.  Would you then have
to use the quadratic formula to solve it?  That seems way too messy.

Date: 05/26/2005 at 23:11:14
From: Doctor Peterson
Subject: re: Factoring and Solving Polynomial Equations

Hi, Kate.

To solve an equation by factoring, the other side has to be zero, so
you don't want to change the right side by adding 3.  In factoring, 
you are paying attention only to the one expression you are factoring,
even if it is part of an equation.

In general, a quadrinomial (four-term polynomial) is very hard to
factor.  So if you are given one, you can expect it to be one you can
do using one of the few available techniques.  They are:

  1. Factor out a common factor (as you tried to do with the x)

  2. Factor by grouping (see below)

  3. Factor three terms as a trinomial, and then combine that with the
     other term as a difference of squares

The last possibility is a rarity.  Your problem can be done by 
grouping.

Then idea is to group the terms into two pairs, and factor out the
greatest common factor in EACH PAIR.  If it works right, the resulting
two factored expressions will then have a common factor that you can
pull out.  Here's an example:

  2x^3 - 3x^2 - 4x + 6

We group the first two terms and the last two terms, and factor out
the GCF from each pair:

  (2x^3 - 3x^2) - (4x - 6)
   x^2(2x - 3)  - 2(2x - 3)

Now we see that the two "terms" both have a factor of 2x-3, so we
factor that out:

  (2x - 3)(x^2 - 2)

If we were lucky, we could now factor that second term (if it were a
difference of squares); since we can't, we're done.

Can you do that with your problem?

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


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/