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How Do You Know When to Stop Factorizing?

Date: 03/09/2010 at 16:32:51
From: Marianne
Subject: Factorize

How do you know when to stop factorizing?

For example,

   6pq - 4qr - 8pr = 2(3pq - 2qr - 4pr)
   
Can 3pq - 2qr - 4pr still be factored?

Or

   10s^2t + 15st^2 + 5s^2 = 5s(2st + 3t^2 + s)

Can (2st + 3t^2 + s) still be factored?

Is there a rule? Thanks.



Date: 03/09/2010 at 23:20:30
From: Doctor Vogler
Subject: Re: Factorize

Hi Marianne,

Thanks for writing to Dr Math. 

That's a good question. The short answer is that there is a way to tell
for certain if you can't factor it anymore, but in practice it's more
complicated than it's worth. Generally, the right answer is that you try
to factor it, and if there isn't an obvious way to factor it, then it
probably can't be factored any more.

If it's a polynomial in only one variable, then see the following for a
discussion of the answer to your question:

http://mathforum.org/library/drmath/view/68494.html

The proper technique for determining whether a polynomial in two or more
variables can be factored or not really belongs to the branch of
mathematics known as Algebraic Geometry. It's pretty advanced subject
matter. But if you know your basic calculus, then I can give you a
simple rule to determine whether a polynomial in two variables can be
factored into a product of two (or more) polynomials, each of degree at
least one. (That is, I'm not counting pulling out an integer factor in
common with the coefficients, as you did in your first example.)

Let's call your (polynomial) function f(x,y). For example, suppose you
want to check if this can be factored:

   f(x,y) = 2xy + 3x^2 + y

If it can, then you would have

   f(x,y) = g(x,y) * h(x,y),

where g and h are polynomials of degree at least one (that is, not
constants). It turns out that there will necessarily be at least one
complex solution (x,y) to the simultaneous equations

   g(x,y) = 0
   h(x,y) = 0

This is known from Bezout's Theorem. (Note that there might not be any
real-number solutions, but there will be at least one complex- number
solution.) Then it turns out that this solution will also be a root of
the two derivatives of f, and of f itself:

   f_x(x,y) = g_x(x,y) * h(x,y) + g(x,y) * h_x(x,y)
   f_y(x,y) = g_y(x,y) * h(x,y) + g(x,y) * h_y(x,y)
   f(x,y) = g(x,y) * h(x,y)

But a typical function in two variables will not have any common roots
of its two derivatives. So if you check, and your function doesn't, then
it can't be factored. (The converse is not necessarily true, but it will
take more work to prove it can't be factored in this case.)

For example, the two derivatives of f(x,y) = 2xy + 3x^2 + y are

   f_x(x,y) = 2y + 6x
   f_y(x,y) = 2x + 1

If both of these are zero, then x = -1/2, and y = 3/2, but that's not 
a solution to f(x,y) = 0, since f(-1/2,3/2) = 3/4.

Similarly, if this can be factored ...

   3pq - 2qr - 4pr

... then so can this:

   f(x,y) = 3xy - 2x - 4y
   
(Can you explain why? Why can't I do that to the other function and 
reduce it to only one variable?)

In particular, this function has

   f_x(x,y) = 3y - 2
   f_y(x,y) = 3x - 4

If both of these are zero, then x = 4/3 and y = 2/3, which is not a
solution to f(x,y) = 0. So this can't be factored either.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to offer
further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/ http://mathforum.org/dr.math/
Associated Topics:
High School Polynomials

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