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Partial fraction expansion

  1. Sep 10, 2015 #1
    1. The problem statement, all variables and given/known data
    divide in partial fractions:

    x^3 -3x^2+x-12 / x^4+5x^2+4

    2. Relevant equations


    3. The attempt at a solution

    I factored x^4+5x^2+4 to (x^2 +1) (x^2+4)

    x^3 -3x^2+x-12/(x^2 +1) (x^2+4) = A (x^2+1)/(x^2 +1) (x^2+4) + B (x^2+4) /(x^2 +1) (x^2+4)

    they all have the same denominator so: x^3-3x^2+x-12 = A (x^2+1) + B (x^2+4)

    now normally my next step would be to try to solve for A by making ( x^2+4) equal to zero so I don't have B in my equation anymore. so I should find an x where x^2 is -4 but this is impossible because of the ^2. how do I solve this? thanks in advance
     
  2. jcsd
  3. Sep 10, 2015 #2

    DEvens

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    You can't use just A and B. You need (ax+b) and (cx+d). You have an x^3 term.
     
  4. Sep 10, 2015 #3

    Ray Vickson

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    You have written
    [tex] x^3 - 3x^2 + x - \frac{12}{x^4} + 5 x^2 + 4 [/tex]
    Are you sure that is what you really want?

    If you mean
    [tex] \frac{x^3 - 3 x^2 + x - 12}{x^4 + 5 x^2 + 4}, [/tex]
    then you absolutely MUST use parentheses, like this: (x^3 -3x^2+x-12) / (x^4+5x^2+4).

    If I were doing it I would expand ##1/(x^4 + 5 x^2 + 4) = 1/(t^2 + 5 t + 4)## into partial fractions in the variable ##t =x^2##, then multiply by the numerator ##x^3 - 3x^2 + x - 12## later. The numerator can be written as
    [tex] x^3 -3 x^2 + x - 12 = (x^3 + x) - (3x^2 + 3) - 9 = (x-3)(x^2+1) - 9 [/tex]
    for the ##(x^2+1)## denominator and as
    [tex] x^3 -3 x^2 + x - 12 = (x^3 + 4x) - (3x^2 + 12) - 3x = (x-3)(x^2 + 4) - 3x [/tex]
    for the ##(x^2+ 4)## denominator; that will allow a complete partial-fraction expansion.

    The only remaining issue is whether the questioner wants/allows partial fractions in ##x^2##, or whether you need to go all the way down to fractions in ##x##. In the latter case your partial fractions will be of the form ##1/(x + i c)##, where ##c## is a real number while ##i = \sqrt{-1}## is the imaginary unit.
     
    Last edited: Sep 10, 2015
  5. Sep 10, 2015 #4

    Mark44

    Staff: Mentor

    No, this isn't right. You have skipped a step here, and your partial fraction decompositions aren't right. Try this to start with:
    ##\frac{x^3 -3x^2+x-12}{(x^2 +1) (x^2+4)} = \frac{Ax + B}{x^2 +1} + \frac{Cx + D}{x^2 +4}##
     
  6. Sep 10, 2015 #5
    how do you guys right fractions and expontents that way? :D would come in handy.

    Would this step be wrong?: (x^3-3x^2+x-12)/[(x^2+1)(x^2+4)] = [(Ax +B) (x^2+1)]/[(x^2 +1) (x^2+4)] + [(Cx+D)(x^2+4)] /[(x^2 +1) (x^2+4)]

    and then say (x^3-3x^2+x-12) = (Ax +B) (x^2+1) + (Bx+c)(x^2+4) becose they all had the same denominator.

    and if so then what is the next step? could turn it into (x^3-3x^2+x-12)= Ax^3 +Bx^2 +Ax +B+ Cx^3 +Dx^2 +4Cx +4D but I don't think that's really useful

    thanks for helping me, was able to solve a lot of the exercises with watching the videos on khanacademy.com on partial fraction expansion but now i'm stuck
     
    Last edited: Sep 10, 2015
  7. Sep 10, 2015 #6

    DEvens

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    To see how somebody has done some equations, click the reply button and look at the tags. Just don't post it because it would be a duplication.

    You wrote (Ax+B) and (Bx+C). You used B twice. You want (Ax+B) and (CX+D).
     
  8. Sep 10, 2015 #7

    vela

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    At this point, you can collect terms on the righthand side, giving you
    $$x^3 - 3x^2 + x - 12 = (A+C)x^3 + (B+D)x^2 + (A+4C) x + (B+4D),$$ and then match coefficients. For example, from the ##x^3## terms, you'd get 1 = A+C. You'll have a system of four linear equations and four unknowns, which is straightforward to solve.
     
  9. Sep 10, 2015 #8
    I found the answer: 1/(x^2+1) - 3/(x^2+4). everybody thanks for the help. I would never been able find the answer without it!
     
  10. Sep 10, 2015 #9

    vela

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    Unfortunately, that's not correct.
     
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