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Re: Factoring
Posted:
Jan 12, 2012 4:48 PM
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On Jan 8, 8:35 pm, "John Smith" <inva...@invalid.invalid> wrote:
> A student asked me how to factor x^3-3x^2-4x+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 x-2. Dividing by x-2 gives x^2-x-6 which is
> easy to factor into (x-3)(x+2) so the answer is (x-2)(x-3)(x+2).
>
> Is there a way to do that whithout guesswork?
>
> Suppose I had a more complicated example such as (2x+1)(3x-2)(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
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