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Partial Fractions problem

  1. Sep 12, 2007 #1
    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
    in the answer they wouldnve had C as 1.

    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. jcsd
  3. Sep 12, 2007 #2
    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,
  4. Sep 12, 2007 #3
    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 [tex](x+1)[/tex] - I used to make that mistake too!

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

    You correctly said that:

    [tex]\frac{4x + 1}{(x+1)^2 (x-2)} = \frac{A}{(x+1)^2} + \frac{B}{(x+1)} + \frac{C}{(x-2)} [/tex]

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

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

    [tex]\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)}[/tex]

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

    [tex]4x+1 \equiv A(x-2)(x+1) + B(x+1)^2(x-2) + C(x+1)^3[/tex] 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
  5. Sep 12, 2007 #4


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    Okay, you know that
    [tex]\frac{4x+1}{(x+1)^2(x-2)}= \frac{A}{x+1}+ \frac{B}{(x+1)^2}+\frac{C}{x-2}[/tex]
    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"
    [tex]\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}[/tex]
    as wished.

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