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Partial Fractions - 3 Unknowns

  1. May 31, 2014 #1
    Hello,

    i've come across a partial fractions problem that I don't know how to solve - Usually, the denominator of the fraction I need to split up into two separate fractions is a quadratic, but in this instance it's a cubic.

    Specifically, the problem i'm having is that two of the factors to this cubic are the same, so when I set x equal to values in order to find the unknowns, everything disappears!

    1. The problem statement, all variables and given/known data

    Write [itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}}[/itex] as the sum of 3 partial fractions.

    2. Relevant equations



    3. The attempt at a solution

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

    [itex] = \frac{A(x + 2)(x + 2)}{(x + 1)(x + 2)(x + 2)} + \frac{B(x + 1)(x + 2)}{(x + 1)(x + 2)(x + 2)} + \frac{C(x + 1)(x + 2)}{(x + 1)(x + 2)(x + 2)}[/itex]

    [itex](x - 1) \equiv A(x +2)(x + 2) + B(x+1)(x + 2) + C(x + 1)(x + 2)[/itex]

    If I set x = -1, then A = -2

    But now if I try to make the other brackets equal to zero, A, B and C disappear...

    I'm probably missing something obvious (or shouldn't be approaching this in the exact same way I would a simpler one), but I can't see the solution.

    Thanks for any help you can give
     
    Last edited: May 31, 2014
  2. jcsd
  3. May 31, 2014 #2

    hilbert2

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    In this case, the correct form of the partial fraction decomposition is ##\frac{x-1}{(x+1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}##. You can calculate the correct decomposition in Wolfram Alpha by entering the command Apart[(x-1)/((x+1)(x+2)^2)]. I hope this helps.
     
  4. May 31, 2014 #3
    I'm not sure I understand why that's the case, but I can see how it helps me solve the question.
     
  5. May 31, 2014 #4

    SteamKing

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    It's not clear what partial fraction decomposition you are trying to write.

    BOAS starts with:

    [itex]\frac{(x - 1)}{(x + 2)(x + 2)^{2}}[/itex]

    Which hilbert2 interprets as:

    [itex]\frac{x-1}{(x+1)(x+2)^2}[/itex]

    and then, the partial fraction decomposition becomes (also, according to hilbert2):

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

    Are you guys solving the same problem?
     
  6. May 31, 2014 #5
    Thank you for catching that.

    In the 'problem statement' section, I did make a typo, but the correct version is used through my working and Hilbert references the correct version.

    I'll edit the OP.

    EDIT 2 Ack, there are more inconsistencies in my post, further edits required.
     
  7. May 31, 2014 #6

    LCKurtz

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    Seems like every post in this thread has a typo or two.

    I assume the second red factor is correct. In that case the right side should be$$
    \frac A {x+1} + \frac B {x+2} + \frac C {(x+2)^2}$$.
     
  8. May 31, 2014 #7
    Ok, I think I have corrected all the typos in my OP.

    I've also answered my own problem;

    [itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}} = \frac{A}{(x + 1)} + \frac{Bx + c}{(x + 2)^{2}}[/itex]

    Letting [itex]C = 2B + D[/itex] Accounts for that extra factor that was confusing me.

    Thanks for the help - I made the same sign error early on in my written working and it cropped up all over the place.
     
  9. Jun 1, 2014 #8

    HallsofIvy

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    There are three unknowns, A, B, and C so you need three equations. Since the given equation is to be true for all x, put in any three values of x you please to get three equations.

    You have
    [tex]\frac{x- 1}{(x+ 1)(x+ 2)^2}= \frac{A}{x+ 1}+ \frac{Bx+ C}{(x+ 2)^2}[/tex]
    Multiply both sides by the denominator to get [itex]x- 1= A(x+ 2)^2+ (Bx+ C)(x+ 1)[/itex]

    Yes, setting x= -1 and -2 will give very simple equations. If x= -1, [itex]-2= A(1)^2+ (B(-1)+ C)(-1+ 1)= A[/itex]. Setting x= -2, [itex]-3= A(-2+ 2)^3+ (B(-2)+ C)(-2+ 1)= 2B- C[/itex].

    So we have A= -2 and 2B- C= -3. Set x equal to what ever third number you like to get a third equation. I happen to thing "0" is a very easy number: if x= 0, -1= A(0+ 2)^2+ (B(0)+ C)(0+ 1)= 4A+ C.

    So we have the three equations, A= -2, 2B- C= -3, and 4A+ C= -1. Since A= -2, that third equation is -8+ C= 1 so C= 9. Then 2B- C= 2B- 9= -3.

    Another way to do problems like this, without setting x equal to any numbers is use the fact that "If two polynomials have the same value for all x, then they must be exactly the same- they must have the same coefficients. We have [itex]x- 1= A(x+ 2)^2+ (Bx+ C)(x+ 1)[/itex]. Multiplying out the right side, [tex]0x^3+ x- 1= A(x^2+ 4x+ 4)+ Bx(x+ 1)+ C(x+ 1)= Ax^2+ 4Ax+ 4A+ Bx^2+ Bx+ Cx+ C= (A+ B)x^2+ (B+ C)x+ (A+C)[/tex] for all x so we must have A+ B= 0, B+ C= 1, and 4A+ C= -1. Solve those three equations for A, B, C.
     
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