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Re: Re: Factoring problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg35536] Re: [mg35515] Re: [mg35499] Factoring problem
*From*: Rob Pratt <rpratt at email.unc.edu>
*Date*: Thu, 18 Jul 2002 03:06:24 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
On Wed, 17 Jul 2002, Rob Pratt wrote:
> On Tue, 16 Jul 2002, Steven Hodgen wrote:
>
> > Hello,
> >
> > I've decided to see if any of you can factor this eq. since it's not
> > possible to have Mathematica show intermediate steps when factoring. This
> > is a problem from my precalc book, and the instructor of the class hasn't
> > been able to get to the final answer either. I've tried and tried using
> > grouping in various ways, as well as other techniques.
> >
> > I'd appreciate it if anyone can figure this out, since I just can't get over
> > this problem.
> >
> > Original problem:
> > y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2
> > Answer in book:
> > (y^2 - p*q)*(y - p)*(y - q)
>
> In[1]:= Expand[(y^2 - p*q)*(y - p)*(y - q)]
>
> 2 2 2 2 3 3 4
> Out[1]= -(p q ) + p q y + p q y - p y - q y + y
>
> So the proposed factorization checks out.
>
> How could you have discovered the factorization? Pretend p and q are
> prime numbers. Then the rational roots theorem would suggest that you try
> the divisors of p^2 q^2 as possible roots. Substitute each of
>
> 1, p, q, p q, p^2 q, p q^2, and p^2 q^2
>
> for y and you will find that both p and q make the expression vanish, so
> y - p and y - q are factors. Dividing these factors out yields y^2 - p q.
>
> Rob Pratt
> Department of Operations Research
> The University of North Carolina at Chapel Hill
>
> rpratt at email.unc.edu
>
> http://www.unc.edu/~rpratt/
I inadvertently omitted p^2 and q^2 from the list of divisors of p^2 q^2
(there should be (2 + 1)*(2 + 1) = 9 total), but these two aren't roots
anyway.
Rob Pratt
Department of Operations Research
The University of North Carolina at Chapel Hill
rpratt at email.unc.edu
http://www.unc.edu/~rpratt/
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