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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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