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May 9th, 2015, 05:24 PM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 673 Thanks: 88  Dividing Polynomials
Can somebody do: (3n^4 + 2n^3  3n^2  2n)/(n + 1)/(n  1) 
May 9th, 2015, 06:10 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
There are two ways of interpreting that. Do you mean ((3n^4 + 2n^3  3n^2  2n)/(n + 1))/(n  1) or (3n^4 + 2n^3  3n^2  2n)/((n + 1)/(n  1)) 
May 9th, 2015, 11:51 PM  #3 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
(3n^4 + 2n^3  3n^2  2n)/(n + 1)/(n  1) = (n^2 * (3n^2 + 2n)  1(3n^2 + 2n)) / (n^21) = (n^21) * (3n^2 + 2n) / (n^21) = 3n^2 + 2n If instead you meant (3n^4 + 2n^3  3n^2  2n)/((n + 1)/(n  1)), as Azzajazz gives as an option, multiply the last result with (n1)^2. 
May 10th, 2015, 06:48 AM  #4 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  $\displaystyle \dfrac{3n^4+2n^3 3n^22n}{(n+1)(n1)} = \dfrac{(3n^43n^2) + (2n^32n)}{n^21} = \\\;\\ = \dfrac{3n^2(n^21)+2n(n^21)}{n^21} = \dfrac{\cancel{(n^21)}(3n^2+2n)}{\cancel{n^21}} = 3n^2+2n$ 
May 10th, 2015, 08:35 AM  #5 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 673 Thanks: 88 
I meant Azzajazz's first way. Isn't that the way it would be done with a normal order of operations?

May 10th, 2015, 08:40 PM  #6  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
What you presented is an abnormal, or deviant, problem of dividing polynomials. The problem would have been presented as, or close to, the first expression that aurel5 wrote.  
May 12th, 2015, 07:38 AM  #7 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
I think the problem as presented is fine. It describes what he wants. We don't have to adapt the formulation of a problem to ones that are written as what one doesn't want.

May 12th, 2015, 03:42 PM  #8  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
I would never give a student a problem in that form. It's deviant. The student was getting bogged down with the interpretation of what to do with the expressions within it. It should have been one polynomial divided by one polynomial in the context of one rational fraction. The problem poser needlessly complicated it for the students. Contrast it with something as the following, where each of A, B, C, D, E, and F are polynomials, is typical (usual): $\displaystyle \dfrac{A}{B} \div \dfrac{C}{D}*\dfrac{E}{F}$  
May 12th, 2015, 07:47 PM  #9 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  

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