Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Factoring Polynomials

Date: 11/07/2001 at 20:28:37
From: Emily
Subject: Factoring

I'm having trouble factoring things like:

12x^3y^9 + 20x^5y^4                 ^ = exponent

and I'm also having trouble FOILING.

Please help - I need it explained better.

Date: 11/08/2001 at 14:51:38
From: Doctor Ian
Subject: Re: Factoring

Hi Emily,

Let's take a look at

  12x^3y^9 + 20x^5y^4

If we break the constants into prime factors, and expand out the 
exponents, we have

  2*2*3*x*x*x*y*y*y*y*y*y*y*y*y  + 2*2*5*x*x*x*x*x*y*y*y*y

Yuck! But now we can identify the pieces that the terms have in 

  2*2*3*x*x*x*y*y*y*y*y*y*y*y*y  + 2*2*5*x*x*x*x*x*y*y*y*y
  ---                              ---

  2*2*3*x*x*x*y*y*y*y*y*y*y*y*y  + 2*2*5*x*x*x*x*x*y*y*y*y
  ---   -----                      ---   -----

  2*2*3*x*x*x*y*y*y*y*y*y*y*y*y  + 2*2*5*x*x*x*x*x*y*y*y*y
  ---   ----- -------              ---   -----     -------

And then we can use the distributive property to move the shared parts 
out front:

  2*2*x*x*x*y*y*y*y(3*y*y*y*y*y + 5*x*x*)

Converting back to exponents, this gives us

  4x^3y^4(3y^5 + 5x^2)

Now, this is kind of messy, so we could do the same thing while 
retaining the exponents:

           12x^3y^9 + 20x^5y^4

         x(12x^2y^9 + 20x^4y^4)         Factor out an x. 

       x^2(12xy^9   + 20x^3y^4)         Again.

       x^3(12y^9    + 20x^2y^4)         Again.

      x^3y(12y^8    + 20x^2y^3)         Factor out a y.

    x^3y^2(12y^7    + 20x^2y^2)         Again.

    x^3y^3(12y^6    + 20x^2y)           Again.

    x^3y^4(12y^5    + 20x^2)            Again.

   2x^3y^4( 6y^5    + 10x^2)            Factor out a 2.

   4x^3y^4( 3y^5    +  5x^2)            Again. 

When you get comfortable doing it the long way, you can start using 
shortcuts.  For example, in this case, we can look at 

           12x^3y^9 + 20x^5y^4

and see that the exponent of x is at least 3 in each term, so we can 
factor out x^3, subtracting 3 from each exponent:

       x^3(12y^9 + 20x^2y^4)

imilarly, we can see that the exponent of y is at least 4 in each 
term, so we can factor out y^4, subtracting 4 from each exponent:

    x^3y^4(12y^5    + 20x^2)            

The greatest common factor of 12 and 20 is 4, so we can factor that 
out, dividing each coefficient by 4:

   4x^3y^4( 3y^5    +  5x^2)            

So this is quicker! But if you try to follow 'factoring rules' without 
understanding what's really going on - i.e., that you're really just 
applying the distributive property over and over until there is 
nothing left to distribute - then you're likely to get mixed up and 
make a mistake. It's better to do things the long way until they get 
so easy that you find yourself looking for a shorter way. 

As for FOILing, don't bother. Use the distributive property instead.  
It makes more sense, and is useful in more situations:

   Distributive Property: (x+2)(x+4)

   When FOIL Fails

Does this help? 

- Doctor Ian, The Math Forum
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum