Confused much, partial fractions

[Wholewheat458]

.. 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:

 

[HallsofIvy]

.. 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.

 

[Wholewheat458]

? How did you pick those number? are they random, or no??
Oh, and can i just say: THANK YOU!

 

[nicksauce]

? 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

 

[Wholewheat458]

^^. thank you, i'll keep it in mind