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  • Maths help- partial fractions
  • Duane…
    Free Member

    Stupid moment here…

    Got to put
    (-11x+80)/((x^2)-16x+64)
    into partial fractions.

    Got
    A/(x-8) + B/(x-8)
    but no idea how to work out A and B?

    Any ideas for a muppet?

    Ta, Duane.

    munkster
    Free Member

    I think the fact that you have a "repeated term" [ie. (x-8)^2] in the denominator changes how you do it but I can't for the life of me work out how it goes.

    ie. I don't think it's simply

    A/(x-8) + B/(x-8)

    but

    A/[(x-8)^2] + B/(x-8)

    but can't remember if it's A or Ax + C on top of your first partial fraction.

    Hopefully *some* help? 😯

    Dobbo
    Full Member

    42

    MS
    Free Member

    Doing an engineering degree, think this is how u do it.

    Its like munkster said

    A/(x-8)^2 + B/(x-8)

    Then you just work out A and B in the normal way

    Cheers

    munkster
    Free Member

    Ah yes, thought I was right but I couldn't get it to work but have just done it…

    So

    A/(x-8)^2 + B/(x-8)

    Multiply second fraction by (x-8) top and bottom

    A/(x-8)^2 + B(x-8)/(x-8)^2

    Write as one fraction and expand/collect numerator

    (A – 8B + Bx)/(x-8)^2

    Equate numerator to numerator of original and compare coefficients/constants

    So B = -11 and A – 8B = 80

    but B = -11 so A + 88 = 80 and A = -8

    Works innit… 😉

    Duane…
    Free Member

    Thanks a ton mate, worked it out after your first post, cheers 🙂

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