# Homework Help: Partial Fractions problem

1. Sep 12, 2007

### Hypochondriac

Im going to Durham uni in oct to do physics, and the nice people of the physics department sent me some maths questions to do before I arrive.

One of the partial fractions questions looked simple enough, but when I did it, I got it wrong....so with the answer they give, i worked back to the question, and they didnt match so I was wondering if it was something I did wrong, or something they did

the question was express (4x+1)/(x+1)^2(x-2) in partial fractions.

so i did the usual 4x+1 ≡ A(x+1)(x-2) + B(x+1)^2(x-2) + C(x+1)^3
then I chose to substitute x=2, as to eliminate A and B,
when I did this I got 9 = 27C, so C = 1/3

Their full answer for the record was:

1/(x+1)^2 - 1/(x+1) + 1/(x-2)

but when substituting 1, -1 and 1 in for A B and C, I get 4x^2 + 4x +1 instead of just 4x+1....

whos wrong?

2. Sep 12, 2007

### Hypochondriac

most of my answer came from memory seen as i gave my text books back to my old school.
and from what i remeber when theres a squared term ie (x+1)^2, in the partial fractions one has the denominator (x+1) and one has (x+1)^2
then i multiplied the letter with that denominator by the denominators it doesn't have.
so the (x+1)^3 came from C(x+1)(x+1)^2, as C had the (x-2) denominator.

which if i was to manually find a common denominator wouldn't make sense,
quess i shouldnt take shortcuts

i know where i went wrong now,
thanks!

3. Sep 12, 2007

### Archduke

I've just done the problem, and I get the same answer as Durham. I think I can see your problem, though. You seem to have multiplied the RHS by an extra $$(x+1)$$ - I used to make that mistake too!

I'll try to show you why you don't need to multiply by the extra $$(x+1)$$.

You correctly said that:

$$\frac{4x + 1}{(x+1)^2 (x-2)} = \frac{A}{(x+1)^2} + \frac{B}{(x+1)} + \frac{C}{(x-2)}$$

What you want to do is make sure the denominators of all of the fractions on the RHS are the same.

So, you need to multiple fraction "B" by $$(x+1)(x-2)$$ so that it has the same denominator as the other two - by that I mean it has a $$(x+1)^2$$ like fraction A, and a $$(x-2)$$ like fraction C. Then, you need to multiply fraction C by $$(x+1)^2$$ so that it contains the same numerator as B, and the same $$(x+1)^2$$ factor as A. Now, to make fraction A the same as the other two, you need to multiply it by $$(x-2)$$. I hope this makes sense.

If we do it your way, we end up with:

$$\frac{4x + 1}{(x+1)^2 (x-2)} = \frac{A(x-2)(x+1) + B(x+1)^2(x-2) + C(x+1)^3}{(x+1)^3 (x-2)}$$

Which is correct (because the x+1 cancel on the RHS), but you can't say that;

$$4x+1 \equiv A(x-2)(x+1) + B(x+1)^2(x-2) + C(x+1)^3$$ because the numerators of the L- and R-HS are not the same.

I seem to have rambled on way too much, but I hope it's helped

Edit: Sorry, I got sidetracked, and ended up being posted way too late!

Last edited: Sep 12, 2007
4. Sep 12, 2007

### HallsofIvy

Okay, you know that
$$\frac{4x+1}{(x+1)^2(x-2)}= \frac{A}{x+1}+ \frac{B}{(x+1)^2}+\frac{C}{x-2}$$
and multiplying by that denominator, (x+1)2(x-2) gives
4x+ 1= A(x+1)(x-2)+ B(x-2)+ C(x+1)2.

No, that doesn't look right, even allowing for differences in what denominator A, B, C had. For one thing, since you only had (x+1)2 in the denominator, you can't have (x+1)3!
Starting from there, you will have, getting "common denominators"
$$\frac{x-2}{(x-2)(x+1)^2}- \frac{(x+1)(x-2)}{(x-2)(x+1)^2}+ \frac{(x+1)^2}{(x-2)(x+1)^2}= \frac{x-2- x^2+ x+ 2+ x^2+ 2x+ 1}{(x-2)(x+1)^2}= \frac{4+1}{(x-2)(x+1)^2}$$
as wished.

Where did you get that (x+1)2 in the second term and and (x+1)3 in the last term?