My Math Forum Factorize completely m^2-2mn+n^2-9r^2

 Algebra Pre-Algebra and Basic Algebra Math Forum

 September 26th, 2014, 12:34 AM #1 Senior Member     Joined: Jan 2012 Posts: 725 Thanks: 7 Factorize completely m^2-2mn+n^2-9r^2 Factorize completely $$m^2-2mn+n^2-9r^2$$ We have $$m(m+2n)+(n-3r)(n+3r)$$ 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: $\displaystyle m^2-2mn+n^2-9r^2=(m-n)^2-(3r)^2$ 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 $\displaystyle m^2-2mn+n^2-9r^2=(m-n)^2-(3r)^2$?
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
Quote:
 Originally Posted by Chikis How is $\displaystyle m^2-2mn+n^2-9r^2=(m-n)^2-(3r)^2$?
$\displaystyle (m-n)^2=(m-n)(m-n)=m^2-mn-mn+n^2=m^2-2mn+n^2$

$\displaystyle (3r)^2=3^2r^2=9r^2$

 September 26th, 2014, 12:06 PM #5 Senior Member     Joined: Jan 2012 Posts: 725 Thanks: 7 \displaystyle \begin{align}m^2-2mn+n^2-9r^2&=(m-n)^2-(3r)^2 \\ &= ((m-n)-3r)((m-n)+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:
 Factor completely: $\:m^2-2mn +n^2-9r^2$

It is a matter of Recognition.

$\quad \underbrace{m^2 - 2mn + n^2}_{\text{square}} - \underbrace{9r^2}_{\text{square}}$

$\quad=\; (m-n)^2 - (3r)^2$

Got it?

September 26th, 2014, 12:55 PM   #7
Newbie

Joined: Sep 2014

Posts: 9
Thanks: 1

Quote:
 Originally Posted by Chikis How is $\displaystyle m^2-2mn+n^2-9r^2=(m-n)^2-(3r)^2$?
m^2-2mn+n^2 is in the form (a-b)^2=a^2-2ab+b^2
got it ....and then use a^2-b^2 formula to find factors

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
Quote:
 Originally Posted by Chikis \displaystyle \begin{align}m^2-2mn+n^2-9r^2&=(m-n)^2-(3r)^2 \\ &= ((m-n)-3r)((m-n)+3r)\end{align} Is that all?
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.

 Tags completely, factorize, m22mn, n29r2

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# m^2-2mn n^2-9r

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ans_now_plz Algebra 7 January 8th, 2013 08:07 AM ThatAnonKid Elementary Math 4 February 8th, 2012 03:24 AM fantom.1040 Algebra 6 July 17th, 2011 12:31 AM naserellid Algebra 14 August 17th, 2010 11:33 AM salamatr25 Algebra 12 August 6th, 2010 06:41 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top