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

Staff Emeritus
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?