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Partial fraction decomposition of the rational expression

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Write the partial fraction decomposition of the rational expression. Check your result algebraically.

    (x2 – 7x + 16)/[(x + 2)(x2 – 4x + 5)]

    3. The attempt at a solution

    [A/(x+2)] + [(Bx+C)/(x2-4x+5)]

    x2-7x+16= A(X2-4x+5)+(Bx+C)(x+2)

    A+B=1 => B=1-A
    -4A+B+C=-7 => Will need to plugin later
    5A+C=16 => C=16-5A

    -4A+B+C=-7 => 4A+1-A+16-5A= -10A=17 A=(-17/10)

    Now I plug A back into the others

    B=1-(-17/10) => B=(27/10)

    C=16-5(-17/10) => C= (49/2)

    Result: [(-17/10)/(x+2)] + [((27x/10)+(49/2))/(x2-4x+5)]

    However, I am being told the answer is wrong. What is it that I am doing wrong?
  2. jcsd
  3. Apr 6, 2010 #2


    Staff: Mentor

    I see a couple of mistakes on the line above, in the coefficients of x and the constant term.
  4. Apr 6, 2010 #3
    I realize there is a mistake, but you could please elaborate upon my mistake without giving me the answer? Your response did not make it obvious or clear for me. Maybe a more indepth response would help me better understand my own mistake. Thank you in advance for taking the time to help me with this problem!

    Is it suppose to be X2-7x+16=[A+B]x2+[-4A+2B+C]x+[5A+2C]

    Is that correct?
  5. Apr 6, 2010 #4


    Staff: Mentor

    That's what I get
  6. Apr 6, 2010 #5
    But now I am stuck because I can only get one value.

    A+B=1 => A=1+B

    -4A+2B+C=-7 =>

    5A+2C=16 =>
  7. Apr 6, 2010 #6


    Staff: Mentor

    Mistake above. A = 1 - B
    You have three equations in three unknowns, so you should be able to solve for them.
  8. Apr 6, 2010 #7
    That was my question. How am I suppose to solve for 3 unknowns when I cannot cancel anything out. I need atleast 2 known variables to solve any of the equations. Do I subtract equations or some sort of method?
  9. Apr 6, 2010 #8
    Use one of the three equations to get a substitution formula. For example, you have A=1-B so far. By substituting into the other two equations, you should be able to reduce your system down to 2 equations and 2 unknowns. Go from there...
  10. Apr 6, 2010 #9
  11. Apr 6, 2010 #10


    Staff: Mentor

  12. Apr 6, 2010 #11
  13. Apr 7, 2010 #12


    Staff: Mentor

    Yes, I'm sure. x2 -4x + 5 can't be factored into linear factors with real coefficients.
  14. Apr 7, 2010 #13
    Oh my god, how stupid of me. I read that wrong.
    I'm so sorry. >.<
  15. Apr 7, 2010 #14


    Staff: Mentor

    We all screw up from time to time...
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