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September 26th, 2014, 12:34 AM   #1
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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?
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September 26th, 2014, 12:42 AM   #2
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I would suggest:

$\displaystyle m^2-2mn+n^2-9r^2=(m-n)^2-(3r)^2$

Now factor as the difference of squares...
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September 26th, 2014, 02:41 AM   #3
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How is $\displaystyle m^2-2mn+n^2-9r^2=(m-n)^2-(3r)^2$?
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September 26th, 2014, 07:50 AM   #4
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Quote:
Originally Posted by Chikis View Post
How is $\displaystyle m^2-2mn+n^2-9r^2=(m-n)^2-(3r)^2$?
$\displaystyle (m-n)^2=(m-n)(m-n)=m^2-mn-mn+n^2=m^2-2mn+n^2$

$\displaystyle (3r)^2=3^2r^2=9r^2$
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September 26th, 2014, 12:06 PM   #5
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$\displaystyle \begin{align}m^2-2mn+n^2-9r^2&=(m-n)^2-(3r)^2 \\
&= ((m-n)-3r)((m-n)+3r)\end{align}$
Is that all?

Last edited by skipjack; September 28th, 2014 at 03:46 AM.
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September 26th, 2014, 12:28 PM   #6
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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?

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September 26th, 2014, 12:55 PM   #7
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Quote:
Originally Posted by Chikis View Post
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
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September 26th, 2014, 01:08 PM   #8
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Quote:
Originally Posted by Chikis View Post
$\displaystyle \begin{align}m^2-2mn+n^2-9r^2&=(m-n)^2-(3r)^2 \\
&= ((m-n)-3r)((m-n)+3r)\end{align}$
Is that all?
That's it.

Last edited by skipjack; September 28th, 2014 at 03:48 AM.
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September 28th, 2014, 03:49 AM   #9
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Or just $(m - n - 3r)(m - n + 3r)$.
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September 28th, 2014, 05:02 AM   #10
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Thank you.
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