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    for 1/(1+x^3),

    I've got to = A/(x+1) + B/(x^2 -x+1)

    But the values of A and B are different depending on x. So they must include an x term? Where do I go from here?
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    (Original post by pippabethan)
    for 1/(1+x^3),

    I've got to = A/(x+1) + B/(x^2 -x+1)

    But the values of A and B are different depending on x. So they must include an x term? Where do I go from here?
    You need to have a polynomial term on top of the fraction which is 1 order lower than the denominator
    EDIT: this is good for most, repeated denominator terms are different though
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    (Original post by pippabethan)
    for 1/(1+x^3),

    I've got to = A/(x+1) + B/(x^2 -x+1)

    But the values of A and B are different depending on x. So they must include an x term? Where do I go from here?
    It should be \displaystyle \frac{1}{1+x^3} = \frac{1}{(x+1)(x^2 - x + 1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 - x + 1}

    If your partal fraction term has polynomial of degree n in the denominator, then the numerator should be of degree n-1. Which is why linear term at the bottom => constant term on top. Quadratic in bottom (like x^2 - x +1) => linear term on top.
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    Aha thank you both! I'll give it another go
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    is this still the case if the denominator is (ax+b)^2 ?
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    (Original post by pippabethan)
    is this still the case if the denominator is (ax+b)^2 ?
    No, if the denominator is (ax+b)^2 then you do \frac{1}{(ax+b)^2} = \frac{A}{ax+b} + \frac{B}{(ax+b)^2}.

    http://tutorial.math.lamar.edu/Class...Fractions.aspx - read this.
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    (Original post by Zacken)
    No, if the denominator is (ax+b)^2 then you do \frac{1}{(ax+b)^2} = \frac{A}{ax+b} + \frac{B}{(ax+b)^2}.

    http://tutorial.math.lamar.edu/Class...Fractions.aspx - read this.
    Thank you
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    (Original post by pippabethan)
    is this still the case if the denominator is (ax+b)^2 ?
    It becomes more complicated. If I had 1/(ax+b)(cx+d)^2 this would be split into A/(ax+b) + B/(cx+d) + (Cx+D)/(cx+d)^2
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    (Original post by EwanWest)
    It becomes more complicated. If I had 1/(ax+b)(cx+d)^2 this would be split into A/(ax+b) + B/(cx+d) + (Cx+D)/(cx+d)^2
    But @Zacken just said this would be C/(cx+d)^2 ??
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    (Original post by pippabethan)
    But @Zacken just said this would be C/(cx+d)^2 ??
    Yes he's correct, I haven't done these in a while. Having said that if you have the extra term in there like I do it will just come out as 0 when you compare coefficients so it won't give you the wrong answer
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    (Original post by pippabethan)
    But @Zacken just said this would be C/(cx+d)^2 ??
    That's correct.
 
 
 
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