Thread: Taylor Series partial fractions

f(x) =
1/
(2x − 5)^2(3x − 1)based at b = 0.
I tried partial fractions and I think I'm doing it wrong.
Thanks in advance.

Originally Posted by khuezy

f(x) =
1/
(2x − 5)^2(3x − 1)based at b = 0.
I tried partial fractions and I think I'm doing it wrong.
[snip]

Then please show your working.

1 = A(2x-5)(3x-1) + B(3x-1) + C(2x-5)(2x-5)
x=5/2
1= B(3(5/2)-1) B=2/13????
That doesn't seem right.

Originally Posted by khuezy

1 = A(2x-5)(3x-1) + B(3x-1) + C(2x-5)(2x-5)
x=5/2
1= B(3(5/2)-1) B=2/13????
That doesn't seem right.

Sorry but I cannot follow this working. Please show all the working, set out clearly. If you expect to get a clear reply to your problem, then surely it's reasonable that you clearly show all your working when requested.