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MathGroup Archive 2002

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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/






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