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### 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.

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
common.

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)
http://mathforum.org/dr.math/problems/lydia.09.18.01.html

When FOIL Fails
http://mathforum.org/dr.math/problems/ryan.03.22.01.html

Does this help?

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

Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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