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September 26th, 2014, 12:34 AM  #1 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7  Factorize completely m^22mn+n^29r^2
Factorize completely $$m^22mn+n^29r^2$$ We have $$m(m+2n)+(n3r)(n+3r)$$ What do I do next? 
September 26th, 2014, 12:42 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs 
I would suggest: $\displaystyle m^22mn+n^29r^2=(mn)^2(3r)^2$ Now factor as the difference of squares... 
September 26th, 2014, 02:41 AM  #3 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7 
How is $\displaystyle m^22mn+n^29r^2=(mn)^2(3r)^2$?

September 26th, 2014, 07:50 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  
September 26th, 2014, 12:06 PM  #5 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7 
$\displaystyle \begin{align}m^22mn+n^29r^2&=(mn)^2(3r)^2 \\ &= ((mn)3r)((mn)+3r)\end{align}$ Is that all? Last edited by skipjack; September 28th, 2014 at 03:46 AM. 
September 26th, 2014, 12:28 PM  #6  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, Chikis! Quote:
It is a matter of Recognition. $\quad \underbrace{m^2  2mn + n^2}_{\text{square}}  \underbrace{9r^2}_{\text{square}} $ $\quad=\; (mn)^2  (3r)^2$ Got it?  
September 26th, 2014, 12:55 PM  #7 
Newbie Joined: Sep 2014 From: hyderabad Posts: 9 Thanks: 1  
September 26th, 2014, 01:08 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra  That's it.
Last edited by skipjack; September 28th, 2014 at 03:48 AM. 
September 28th, 2014, 03:49 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1980 
Or just $(m  n  3r)(m  n + 3r)$.

September 28th, 2014, 05:02 AM  #10 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7 
Thank you.


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