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May 9th, 2015, 05:24 PM   #1
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Dividing Polynomials

Can somebody do:

(3n^4 + 2n^3 - 3n^2 - 2n)/(n + 1)/(n - 1)
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May 9th, 2015, 06:10 PM   #2
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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))
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May 9th, 2015, 11:51 PM   #3
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(3n^4 + 2n^3 - 3n^2 - 2n)/(n + 1)/(n - 1) =
(n^2 * (3n^2 + 2n) - 1(3n^2 + 2n)) / (n^2-1) =
(n^2-1) * (3n^2 + 2n) / (n^2-1) = 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 (n-1)^2.
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May 10th, 2015, 06:48 AM   #4
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$\displaystyle \dfrac{3n^4+2n^3 -3n^2-2n}{(n+1)(n-1)} = \dfrac{(3n^4-3n^2) + (2n^3-2n)}{n^2-1} =
\\\;\\
= \dfrac{3n^2(n^2-1)+2n(n^2-1)}{n^2-1} = \dfrac{\cancel{(n^2-1)}(3n^2+2n)}{\cancel{n^2-1}} = 3n^2+2n$
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May 10th, 2015, 08:35 AM   #5
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I meant Azzajazz's first way. Isn't that the way it would be done with a normal order of operations?
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May 10th, 2015, 08:40 PM   #6
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Quote:
Originally Posted by EvanJ View Post
I meant Azzajazz's first way. Isn't that the way it would be done with a normal order of operations?
Yes, but you didn't present a "normal" problem of dividing polynomials.

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.
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May 12th, 2015, 07:38 AM   #7
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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.
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May 12th, 2015, 03:42 PM   #8
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Quote:
Originally Posted by Hoempa View Post
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.
"fine?"

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):

$\displaystyle \dfrac{A}{B} \div \dfrac{C}{D}*\dfrac{E}{F}$
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May 12th, 2015, 07:47 PM   #9
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Quote:
Originally Posted by EvanJ View Post
Can somebody do:

(3n^4 + 2n^3 - 3n^2 - 2n)/(n + 1)/(n - 1)

Could be a typo ( somewhere ).
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