
Re: Factoring
Posted:
Jan 12, 2012 4:48 PM


On Jan 8, 8:35 pm, "John Smith" <inva...@invalid.invalid> wrote: > A student asked me how to factor x^33x^24x+12 > So I took a guess (correctly it seems) that since 12 factors as 2,2,3 one of > the factors was likely to be x2. Dividing by x2 gives x^2x6 which is > easy to factor into (x3)(x+2) so the answer is (x2)(x3)(x+2). > > Is there a way to do that whithout guesswork? > > Suppose I had a more complicated example such as (2x+1)(3x2)(4x+5) I > haven't multiplied it out but how would I find the factors if I didn't know > them in advance? > > Thanks for any enlightenment. > > old guy
I see you've received many answers, and one of them may contain what I'm about to write; I don't have time to look through them all and see.
Here are some basic ideas you can try to factor a polynomial f with integer coefficients, into pieces with integer coefficients:
1. You can look for rational roots. (Note that this is sufficient to get a factor, but often not necessary.) There's a well known theorem that such a root must be of the form + a/b, where a divides the constant term of f and b divides the leading coefficient of f. In your case, the root must be + some factor of 12. Just try them.
2. It's easy to see that any such factor must have leading coefficient dividing the leading coefficient of f, and constant term dividing that of f. Since your f is cubic, at least one term must be linear. This reduces to the same thing as (1) in this case.
3. Variations of the idea in (2) can often be used to find interesting factors. For example, try factoring x^4 + 1 over R, not Z. Obviously there is no linear term. So the only hope is to find two quadratics. From (2), it's reasonable to try x^2 + a*x + 1, x^2 + b*x + 1; or maybe use 1 instead of the +1. This leads to a simple little system of equations that can be solved for a and b.
Robert H. Lewis Fordham University

