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Solving a Perfect Square Trinomial


Date: 05/11/98 at 14:42:09
From: Joe
Subject: Solutions for perfect square trinomial

How do I solve this kind of problem?

   m^2 - 4m + 4 - n^2 + 6n - 9

This is as far as I have gotten:

   (m^2 - 4m + 4) - (n^2 - 6n + 9)
   (       )^2 - (      )^2

Please help.


Date: 05/11/98 at 17:04:51
From: Doctor Sam
Subject: Re: Solutions for perfect square trinomial

Joe,

You began by grouping together the terms that sort of seem to go 
together:

   (m^2 - 4m + 4) - (n^2 - 6n + 9)

The next step is to notice that the trinomials in each parentheses are 
perfect squares. That is:

   m^2 - 4m + 4 = (m-2)^2    and 
   n^2 - 6n + 9 = (n-3)^2

If you don't see this you can always try factoring the expressions 
using trinomial factoring: 

   m^2 - 4m + 4 = (m - ?)(m -?) 

and you will see that both factors must be the same.

So now we have changed the original problem into (m-2)^2 - (n-3)^2.  
What next?  Well, I cannot help but notice that this is a difference 
of squares. 
 
I assume that you are familiar with the way something like A^2 - B^2 
factors?  It is a difference of two squares, and its factors are:

   (A - B)(A + B)

This is a pattern that you can use NO MATTER WHAT A and B stand for.

In your problem A = m - 2 and B = n - 3. So: 
  
   (n-2)^2 - (m-3)^2 = ([m-2]-[n-3]) ([m-2]+[n-3])

                     =  (m - n + 1) (m + n - 5)

I hope that helps.

-Doctor Sam,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
Middle School Factoring Expressions

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