Partial fraction decomposition

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Homework Help Overview

The discussion revolves around the topic of partial fraction decomposition, specifically for the expression (7x^3 - 2)/[(x^2)(x+1)^3]. Participants are exploring how to correctly set up the decomposition given the structure of the denominator.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct form of the partial fraction decomposition, questioning the number of terms needed based on the powers of the factors in the denominator. There is also exploration of alternative setups for the decomposition.

Discussion Status

The conversation is active, with participants providing insights into the necessary components of the decomposition. Some guidance has been offered regarding the structure of the decomposition, but there is no explicit consensus on a single approach yet.

Contextual Notes

Participants are navigating the rules of partial fraction decomposition, particularly concerning the treatment of repeated factors in the denominator. There is an indication of uncertainty regarding the setup of the fractions.

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can someone help me set up this problem. it asks for the partial fraction decomposition of:

(7x^3 - 2)/[(x^2)(x+1)^3]

i thought you put A/x^2 + B/(x+1)^3 and solve but it doesn't work that way.
 
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It's not that simple: when your factors in the denominator are raised to a power n, you need n instances of that factor in your decomposition proposal (one for each exponent, 1 -> n). In your case:

[tex] \frac{{7x^3 - 2}}{{x^2 \left( {x + 1} \right)^3 }} = \frac{A}{{x^2 }} + \frac{B}{x} + \frac{C}{{\left( {x + 1} \right)^3 }} + \frac{D}{{\left( {x + 1} \right)^2 }} + \frac{E}{{x + 1}}[/tex]

Do you get the idea?
 
yeah that makes sense. is there another way to do it where you put Ax+B over the x^2 and Cx^2 + Dx + E over the (x+1)^3 and do it with just two fractions?
 
Sure, since (my A and B are the other way arround though)

[tex]\frac{A}{{x^2 }} + \frac{B}{x} = \frac{A}{{x^2 }} + \frac{{Bx}}{{x^2 }} = \frac{{A + Bx}}{{x^2 }}[/tex]

I don't see how this would make any significant difference though...
 
ok cool - I am sure it wouldn't but i just wanted to be sure.
 

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