RE: Re: Factoring problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg35534] RE: [mg35515] Re: [mg35499] Factoring problem*From*: "DrBob" <majort at cox-internet.com>*Date*: Thu, 18 Jul 2002 03:06:20 -0400 (EDT)*Reply-to*: <drbob at bigfoot.com>*Sender*: owner-wri-mathgroup at wolfram.com

Much easier than my solution! Thanks. Bobby -----Original Message----- From: Rob Pratt [mailto:rpratt at email.unc.edu] Subject: [mg35534] [mg35515] Re: [mg35499] Factoring problem 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/