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/