Confused much, partial fractions

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Discussion Overview

The discussion revolves around the method of partial fractions, specifically how to determine constants in the decomposition of rational functions. Participants explore the application of this method to specific equations and share techniques for solving them.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about finding three constants in the partial fraction decomposition of a rational function.
  • Another participant suggests simplifying the problem by multiplying both sides of the equation by the common denominator to eliminate the fractions.
  • It is proposed that participants can either equate coefficients after expanding or substitute specific values of x to create equations for solving the constants.
  • Specific values of x (-2, 1, and 2) are recommended for their utility in simplifying the equations.
  • Another participant questions the choice of these specific values, seeking clarification on whether they were randomly selected.
  • Appreciation is expressed for the advice given regarding the method of choosing values that simplify the problem.

Areas of Agreement / Disagreement

Participants generally agree on the method of using specific values to simplify the process of finding constants, though there is some uncertainty about the selection of those values.

Contextual Notes

The discussion does not resolve the underlying confusion about the initial problem or the implications of having a numerator without a variable. There are also no explicit definitions provided for the terms used in the equations.

Who May Find This Useful

Students or individuals learning about partial fraction decomposition in algebra or calculus may find this discussion helpful.

Wholewheat458
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.. oy, I'm just not sure how to find 3 constants!
Here is my problem:
5x^2-4/(x-2)(x+2)(x-1) = A/(x-2)+B/(x+2)+C/(x-1)
.. i got a bit of it done, but it's all wrong
OH! and what am i supposed to do if the numerator of the first equation does not have any sort of variable with it??
my next problem:
3/r^2-4r-5 = A/r-5 + B/r+1
Well, any help would be much appreciated ^^ :shy:
 
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Wholewheat458 said:
.. oy, I'm just not sure how to find 3 constants!
Here is my problem:
5x^2-4/(x-2)(x+2)(x-1) = A/(x-2)+B/(x+2)+C/(x-1)
.. i got a bit of it done, but it's all wrong
OH! and what am i supposed to do if the numerator of the first equation does not have any sort of variable with it??
my next problem:
3/r^2-4r-5 = A/r-5 + B/r+1
Well, any help would be much appreciated ^^ :shy:
First, you can considerably simplify the problem by multiplying both sides of the equation by that common denominator:
5x^2- 4= A(x+2)(x-1)+ B(x-2)(x-1)+ C(x-2)(x+2)

Now you can do any of several things:

You could multiply out the left side and equate "corresponding" coefficients.

Or, since that is true for all x, you could just pick 3 values of x and get 3 equations to solve for A, B, and C.

In particular, you will find taking x= -2, x= 1, and x= 2 are especially nice!

For 3/r^2-4r-5 = A/r-5 + B/r+1, the same advice: multiply both sides by (r-5)(r+ 1) and you get 3= A(r+1)+ B(r-5). Now let r= -1 and r= 5.
 
? How did you pick those number? are they random, or no??
Oh, and can i just say: THANK YOU!
 
Wholewheat458 said:
? How did you pick those number? are they random, or no??
Oh, and can i just say: THANK YOU!

If you mean by choosing to test with x = -2, x = 1, and x=2, those numbers were clearly chosen so that all but one of the terms goes to zero. I must also add on that this is a remarkably good trick!Cheers,
Nick
 
^^. thank you, i'll keep it in mind
 

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