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Posted by on Sunday, February 24, 2008 at 3:25pm.

My Calculus class just started Partial Fractions, and I understand MOST of it.

I'm having a slight problem though, when either C, or C and D come in.

I understand

A/(something) + B/(something)

But I become confused when it's either

A/(something) + B/(something) + C/(something)


A/(something) + B/(something) + (Cx + D)/(something)

Why is it that you get ONLY C, or Cx + D ?

I always get confused when it comes to that, so help would be greatly appreciated.

  • Calculus (Partial Fractions) - , Sunday, February 24, 2008 at 4:23pm

    When you have a quadratic that does not have real roots in the denominator (complex roots only), you put Cx+D in the numerator.
    like if your denominator were (x-1)(x+1)(x-2)
    you might use
    A/(x-1) + B/(x+1) +C/(x-2)
    But if your denominator were:
    You could deal with the (x-1) part just fine.
    but what to do with the other part?
    You can not write it as the sum of two fractions with numerators B and C because you can nor write
    x^2-2x+2 as (x+p)(x+q)
    so you have to resort to
    (C + D x)/(x^2-2x+2)

  • Calculus (Partial Fractions) - , Sunday, February 24, 2008 at 4:42pm

    This method has practical applications only if the denominator factors.
    Since you don't give an example I will supply one

    separate (5x^2+3x+4)/(x^3+x^2-2x)

    (I started with known fractions and simplified, so that I would have a question that worked out)

    The bottom factors to x(x+2)(x-1)

    so let
    (5x^2+3x+4)/(x^3+x^2-2x) = A/(x+2) + B/(x-1) + c/x

    = [Ax(x-1) + Bx(x+2) + C(x+2)(x-1)]/x(x+2)(x-1)

    clearing the denominator we get:
    5x^2+3x+4 = Ax(x-1) + Bx(x+2) + C(x+2)(x-1)

    now let x=0, then -2C = 4, and C = -2
    let x=1, then 3B=12, and B = 4
    let x=-2, then 6A = 18 and A = 3

    so my original fraction (5x^2+3x+4)/(x^3+x^2-2x) can
    be split into
    3/(x+2) + 4/(x-1) - 2/x

  • Calculus (Partial Fractions) - , Monday, February 25, 2008 at 2:28pm

    Reiny. You misunderstood. And I didn't word it right.

    I meant where does (Cx + D)/Something

    Pop up?

  • Calculus (Partial Fractions) - , Monday, February 25, 2008 at 3:11pm

    Okay. Example.

    If the Denominator is

    x^4 - 2x^2 - 8 > (x-2) (x+2) (x^2+2)

    Would I do A + B + C, or A + B + (Cx + D)?

  • Calculus (Partial Fractions) - , Monday, February 25, 2008 at 3:15pm

    Wow. I'm making this topic last a while.

    I think- and this is just out loud- that you would do (Cx + D) Over the X^2 + 2?

    Am I getting that right?


    You have Cx + D over something if the SOMETHING has a term of X to a power greater than one?

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