Partial Fractions: Reducing x^5, Example Help Needed

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The discussion focuses on reducing the expression \(\frac{x^5}{x^3 - x}\) using polynomial long division and factoring techniques. Participants clarify that the numerator can be simplified by first dividing both the numerator and denominator by 'x', resulting in \(\frac{x^4}{x^2 - 1}\). Further simplification involves recognizing that \(x^4\) can be expressed as \(x^2(x^2 - 1) + x^2\), leading to the final form \(x^2 + \frac{x^2}{x^2 - 1}\). This method provides a clear path for tackling similar polynomial fraction problems.

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I`ve been asked to complete the following expresion
[tex]\frac{x^5}{x^3 - x}[/tex]

I know i`m supossed to reduce the numerator, but i`m a little stuck getting started.

The problem i have is how do you reduce [tex]x^5[/tex]?

is it simply [tex]x^3(x^2)[/tex]? but then how do you get rid of the [tex]x^3[/tex]

AAHH i`m lost.

I obviously don`t want you to give me the answer so an example of your choice would be greatly appreciated.

Cheers
 
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You can first divide both denominator and numerator by 'x'. They have 'x' in common. So:
[tex]\frac{x ^ 5}{x ^ 3 - x} = \frac{x ^ 4}{x ^ 2 - 1}[/tex]
You can use 'polynomial long division' to reduce the degree of the numerator. You can click here for more information.
-------------------
Or you can also do it a little bit differently:
You notice that x4 = (x2)(x2). But the denominator is x2 - 1, so:
x4 = x2(x2 - 1) + x2.
Now:
[tex]\frac{x ^ 4}{x ^ 2 - 1} = \frac{x ^ 2(x ^ 2 - 1) + x ^ 2}{x ^ 2 - 1} = x ^ 2 + \frac{x ^ 2}{x ^ 2 - 1}[/tex]
Now just do the same for [tex]\frac{x ^ 2}{x ^ 2 - 1}[/tex]:
[tex]\frac{x ^ 2}{x ^ 2 - 1} = \frac{x ^ 2 + ... - ...}{x ^ 2 - 1} = ...[/tex]
Viet Dao,
 

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