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    im gonna sound like an idiot but my answers never turn out for questions when the denominator is JUST a ^x term eg.
    2x/(x+2)^2

    i can do it when its say 2x/(x-2)(x+2)^2

    but not just when theres a double root by its self!! PLease help
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    (Original post by kashif_k)
    im gonna sound like an idiot but my answers never turn out for questions when the denominator is JUST a ^x term eg.
    2x/(x+2)^2

    i can do it when its say 2x/(x-2)(x+2)^2

    but not just when theres a double root by its self!! PLease help
    \frac{2x}{(x+2)^{2}}=\frac{A}{(x  +2)}+\frac{B}{(x+2)^{2}}
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    (Original post by Gaz031)
    \frac{2x}{(x+2)^{2}}=\frac{A}{(x  +2)}+\frac{B}{(x+2)^{2}}

    i know i gotta do that. But i substitute radom number in the form to equations and then take them away form each other abolishing either A or B. but my answer never turns out to be correct. Am i doing something wrong?
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    (Original post by kashif_k)
    i know i gotta do that. But i substitute radom number in the form to equations and then take them away form each other abolishing either A or B. but my answer never turns out to be correct. Am i doing something wrong?
    Multiply everything by (X+2)², cancel down, then equate the co-efficients.

    2x = Ax + 2A + B

    A = 2

    2(2) + B = 0

    B = -4

    I think that's right.
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    (Original post by kashif_k)
    im gonna sound like an idiot but my answers never turn out for questions when the denominator is JUST a ^x term eg.
    2x/(x+2)^2
    \frac{2x}{(x+2)^{2}} \equiv \frac{A}{(x+2)}+\frac{B}{(x+2)^{  2}}\\

2x \equiv A(x+2)+B\\

x=-2 \rightarrow B=-4\\

x=0 \rightarrow A=2\\

\frac{2x}{(x+2)^{2}} \equiv \frac{2}{(x+2)}-\frac{4}{(x+2)^{2}}\\

\textit{You can check that this answer is correct by forming a common denominator between the two terms}
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    oooopppps!! just realised what it is i do wrong!! im such a plum, i forgot that i dont need to B/(x+2)^2 by anything, what i was doing was this

    2x =A(x+2)^2 + B(x+2)

    DOH!!!

    thank for helping me out
 
 
 
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