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Math Help - Express 5x^2+7x-9/4-8x+x^2-2x^3 in partial fraction

  1. #1
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    Express 5x^2+7x-9/4-8x+x^2-2x^3 in partial fraction

    Hi guys,

    as stated above, express 5x^2+7x-9/4-8x + x^2-2x^3 in partial fraction

    My workings:

    5x^2+7x-9/4-8x+x^2-2x^3 = 5x^2+7x-9/4(1-2x)+x^2(1-2x)
    = 5x^2+7x-9/(4+x^2)(1-2x)
    = A/(1-2x) + (Bx+C)/(4+x^2)
    = A(4+x^2)+(Bx+C)(1-2x)/4-8x+x^2-2x^3
    5x^2+7x-9 = A(4+x^2)+(Bx+C)(1-2x)

    Sub x=1/2
    5(1/2)^2+7(1/2)-9=A(4+(1/2)^2)
    -17/4=(17/4)A
    A=-1

    Sub x=0
    -9=-1(4)+(C)(1)
    C=-5

    Sub x=1
    5(1)^2+7(1)-9=-1(5)+(B-5)(-1)
    3=-5-B+5
    B=-3

    Answer: 5x^2+7x-9/4-8x+x^2-2x^3 = -1/(1-2x) - 3x+5/(x^2+4)

    However, the answer from me assessment book is -(1/(2x-1)-(3x+5)/(x^2+4)

    What had i done wrongly?

    Thanks a lot
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  2. #2
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    Re: Express 5x^2+7x-9/4-8x+x^2-2x^3 in partial fraction

    \frac{5x^2+7x-9}{4-8x + x^2-2x^3}

    \frac{5x^2+7x-9}{4(1-2x) + x^2(1-2x)}

    \frac{5x^2+7x-9}{(4+x^2)(1-2x)}


    \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{A}{1-2x} + \frac{Bx+C}{x^2+4}


    5x^2+7x-9 = A(x^2+4) + (Bx+C)(1-2x)

    5x^2+7x-9 = Ax^2+4A + Bx - 2Bx^2 +C - 2Cx

    5x^2+7x-9 = (A-2B)x^2 + (B-2C)x + (4A+C)

    equating coefficients ...

    A-2B = 5

    B-2C = 7

    4A+C = -9

    solving the system of equations ...

    A = -1 \,  , \, B = -3 \, , \, C = -5

    \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{1}{2x-1} - \frac{3x+5}{x^2+4}

    Answer: 5x^2+7x-9/4-8x+x^2-2x^3 = -1/(1-2x) - 3x+5/(x^2+4)

    However, the answer from me assessment book is -(1/(2x-1)-(3x+5)/(x^2+4)
    I agree with this solution.
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  3. #3
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    Re: Express 5x^2+7x-9/4-8x+x^2-2x^3 in partial fraction

    Quote Originally Posted by skeeter View Post
    \frac{5x^2+7x-9}{4-8x + x^2-2x^3}

    \frac{5x^2+7x-9}{4(1-2x) + x^2(1-2x)}

    \frac{5x^2+7x-9}{(4+x^2)(1-2x)}


    \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{A}{1-2x} + \frac{Bx+C}{x^2+4}


    5x^2+7x-9 = A(x^2+4) + (Bx+C)(1-2x)

    5x^2+7x-9 = Ax^2+4A + Bx - 2Bx^2 +C - 2Cx

    5x^2+7x-9 = (A-2B)x^2 + (B-2C)x + (4A+C)

    equating coefficients ...

    A-2B = 5

    B-2C = 7

    4A+C = -9

    solving the system of equations ...

    A = -1 \,  , \, B = -3 \, , \, C = -5

    \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{1}{2x-1} - \frac{3x+5}{x^2+4}



    I agree with this solution.
    Thank you so much.

    I'm assuming that the answer from my assessment book is wrong, am I right? I can't seems to find any way to achieve that answer
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