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resolve the partial fraction:x^3+6/x^2(x+3).with explanations thanks

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  1. First, since the degree of the top is at least as big as that of the bottom, just do a long division, and you get

    (x^3+6)/(x^2 (x+3)) = 1 + (-3x^2+6)/(x^2 (x+3))

    Now you want the partial fractions for the remainder, in the form

    (-3x^2+6)/(x^2 (x+3)) = A/x + B/x^2 + C/(x+3)
    =
    Ax(x+3)+B(x+3)+Cx^2
    ------------------------
    x^2(x+3)
    =
    Ax^2+3Ax+Bx+3B+Cx^2
    ----------------------
    x^2(x+3)
    =
    (A+C)x^2 + (3A+B)x + 3B
    -------------------------
    x^2(x+3)

    In order for those two fractions to be identical, all the coefficients of all the powers of x must match. That means we have

    A+C = -3
    3A+B = 0
    3B = 6

    That's easy to solve, and we get
    A = -2/3
    B=2
    C = -7/3

    so our partial fraction for the complete original fraction is

    1 + (-2/3)/x + 2/x^2 + (-7/3)/(x+3)

    or 1 - 2/(3x) + 2/x^2 - 7/(3(x+3))

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  2. You fixed Damon's concern, but still brackets are missing
    I will assume you meant:

    (x^3+6)/(x^2(x+3) )

    I first did a long algebraic division to get it to
    1 - (3x^2 - 6)/(x^2(x+3) )

    So let's just work on that last term

    let (3x^2 - 6)/(x^2(x+3) ) = A/x + (Bx + C)/x^2 + D/(x+3)
    = ( Ax(x+3) + (x+3)(Bx + C) + Dx^2)/(x^2(x+3))

    then 3x^2 - 6 = Ax(x+3) + (x+3)(Bx + C) + Dx^2
    let x = 0 , -6 = 0 + 3C + 0
    C = -2
    let x = -3, 21 = 0 + 0 + 9D
    D = 21/9 = 7/3
    let x = 1 , -3 = 4A + 4(B - 2) + D = -3
    4A + 4B - 8 + 7/3 = -3
    A + B = 2/3

    let x = 3 --> 18A + 6(3B -2) + 9D = 21
    18A + 18B -12 + 63/3 = 21
    18A + 18B = 12
    A+B = 2/3

    no matter what I try for x, I get A + B = 2/3
    So we can let A and B be anything, why not make it simple and let B = 0, then A = 2/3

    then :
    (3x^2 - 6)/(x^2(x+3) ) = (2/3) / x + (-2)/x^2 + (7/3) / (x+3)

    = 2/(3x) - 2/x^2 + 7/(3(x+3))

    finally:

    (x^3+6)/(x^2(x+3) ) = 1 - (3x^2 - 6)/(x^2(x+3))
    = 1 - (2/(3x) - 2/x^2 + 7/(3(x+3)) )

    = 1 - 2/(3x) + 2/x^2 - 7/(3(x+3))

    Wheuhhh!

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  3. anonymous made the "lucky" assumption that the second fraction ?/x^2 had a constant for a numerator.
    In general you cannot make such an assumption.
    If the denominator is a second degree expression, like x^2, then the numerator could be a first degree expression of the form Bx + k

    notice the values of A,B, C, and D are not defined the same way as those by anonymous.

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  4. tank you reiny

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