# Homework Help: Partial fractions

1. Jan 25, 2005

### whkoh

Given

$$\frac{2+5x+15x^2}{\left (2-x\right )\left (1+2x^2\right )}=\frac{8}{2-x} + \frac{x-3}{1+2x^2}$$

I am asked to deduce the partial fractions of:

$$\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}$$

I can solve it using my usual method, but that's not what the question requires. Any help? I can't see the link between the two.

2. Jan 25, 2005

### vincentchan

$$\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}=\frac{A}{1-x} + \frac{B}{1+8x^2}$$

and solve for A and B... this is the standard approach, is my method UNUSUAL? :surprised

3. Jan 25, 2005

### vincentchan

i have a not so usual method... substitude x=2u in your first equation... and you will get the second one. is this what you looking for?

4. Jan 25, 2005

### whkoh

Yes, that's my usual way of solving such questions. But I still don't get it...

5. Jan 25, 2005

### marlon

This is wrong
it should be :
$$\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}=\frac{A}{1-x} + \frac{Bx +C}{1+8x^2}$$

Now determin A,B and C by adding up these fractions and then compare the numerator with the given fraction...

marlon

Last edited: Jan 25, 2005
6. Jan 25, 2005

### whkoh

Marlon: thanks for your help, but the question specified that I deduce and not solve.

Vincent: substituting x=2u works. Thanks.

7. Jan 25, 2005

### marlon

i know, but that answer is already given. I just wanted to point out that the given partial fractions were wrong,...that's all

marlon

8. Jan 25, 2005

### vincentchan

me wrong?
did I say B is constant?

9. Jan 25, 2005

### dextercioby

Compute A and B with your formula...

Daniel.