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Math Help - Paratial fractions-need help

  1. #1
    Junior Member
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    Paratial fractions-need help

    I've worked some problems, but I keep on getting stuck. Can you please help me?

    Write the partial fraction decomposition of the rational expression.

    1. (7x^2-x-12)/(x)(x+1)(x-1)

    Here is what I did:

    (x)(x+1)(x-1)[(7x^2-x-12)/(x)(x+1)(x-1)] = (x)(x+1)(x-1)[(a)/(x) + (b)/(x+1) + (c)/(x-1)]

    Then I find the terms that can cancel out and I get:

    7x^2-x-12= a(x+1)(x-1) + b(x-1)x+ c(x+1)x
    7x^2-x-12=ax^2-a+bx-b+bx+cx+c+cx
    7x^2-x-12= ax^2-a+bx^2-b+cx^2+c

    Then I get the first equation: a+b=7, this is where I get stuck.

    3. (9x^2-x-14)/(x^3-x)

    (x)(x+1)(x-1)[(9x^2-x-14)/(x)(x+1)(x-1)= (x)(x+1)(x-1)[ (a)/(x) + (b)/(x+1)+ (c)/(x-1)

    9x^2-x-14= a(x+1)(x-1)+ b(x-1)x + c(x+1)x
    9x^2-x-14= ax^2-a-bx^2-b+cx^2+c
    9x^2-x-14= ax^2-bx^2+cx^2-a-b+c

    The equations:
    a-b+c=9
    b=0
    b+c=-1
    c=-14


    Write the partial fraction decomposition of the rational expression.

    11. (6x+5)/((x-9)^2)

    (x-9)^2[(6x+5)/((x-9)^2)]= (x-9)^2[(a)/(x-9) + (b)/((x-9)^2))

    (6x+5)/(x-9)(a+b)
    (6x+5)/ax+bx-9a-9b

    the equations:

    a+b=6
    -9a-9b=5

    To get a variable to cancel, I mulitplied the first equation by 9 and got:

    9a+9b=54
    -9a-9b=5

    both 9a and 9b cancel out and I just 59.


    13. (5x^2-5x+7)/((x-1)^3)

    ((x-1)^3)(5x^2-5x+7)/((x-1)^3)= ((x-1)^3))[ (a)/(x-1) + (b)/((x-1)^2)+ (c)/((x-1)^3)

    (5x^2-5x+7)=((x-1)^2)(a) + (x-1)(b) +c
    (5x^2-5x+7)=ax^2+a+bx-b+c

    The equations I couldnt get on this problem.

    15. (x+4)/(x^3-2x^2+x)- This one I dont understand.

    Write the partial fraction decomposition of the rational expression.

    17. (12x+3)/ (x^3-1) - This one I dont understand

    20. (12x^2-12x+6)/(x-3)(x^2+4)

    6(2x^2-2x+1)=(x-3)(x^2+4) [(a)/(x-3) + (bx+c)/(x^2+4)
    6(2x^2-2x+1)= a(x^2+4) + bx+c(x-3)
    6(2x^2-2x+1) = ax^2+4a+bx^2-3bx+cx-3c

    The equations:
    a+b=6
    4a-3b (this is where I get stuck)

    Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants.

    26. (7x-3)/((x^2+x-7)^2)

    I got: (ax+b)/(x^2+x-7) + (cx+d)/((x^2+x+7)^2)

    29. (2x+1)/(x-6)((x^2+x-4)^2)

    I got: (a)/(x-6) + (bx+c)/(x^2+x-4)

    Write the partial fraction decomposition of the rational expression.


    31. (x^2+4x-1)/((x^2+3)^2)

    x^2+4x-1 = ax +b(x^2+3)+cx+b/((x^2+3)^2)
    x^2+4x-1= ax+b(x^2+3)+cx+d
    x^2+4x-1=ax^3+3a+bx^2+3b+cx+d
    x^2+4x-1=ax^3+bx^2+ (3a+c)x + 3b+d

    The equations:
    a=1
    b=0
    3a+c=4
    3b+d=-1

    When I come up with the fraction I get: (x+4)/(x^2+3) + (-4x - 12)/((x^2+3)^2))

    33. (4x^3+ 4x^2)/((x^2+5)^2))

    I did: (4x^3+ 4x^2)= x^2+5 (ax+b) + cx+d
    (4x^3+ 4x^2)= ax^3+bx^2+ 5a+5b+cx+d

    The equations:
    a=4
    b=4
    5a+c=0
    5b+d=0

    The final answer I got: (4x+4)/(x^2+1) + (-20x-20)/((x^2+5)^2)
    Last edited by soly_sol; November 25th 2007 at 03:27 PM. Reason: Forgot to show my work
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by soly_sol View Post
    I've worked some problems, but I keep on getting stuck. Can you please help me?

    Write the partial fraction decomposition of the rational expression.

    1. (7x^2-x-12)/(x)(x+1)(x-1)
    Put

     <br />
\frac{7x^2-x-12}{x(x+1)(x-1)} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}<br />

    .......... <br />
=\frac{A(x^2-1)+Bx(x+1)+Cx(x-1)}{x(x-1)(x+1)}<br /> <br />
=\frac{x^2(A+B+C)+x(B-C)+(-A)}{x(x-1)(x+1)}<br />

    So equating coefficients in the denominators we are left with the system of equations:

    A+B+C = 7

    B-C=-1

    -A=-12

    to solve.

    RonL
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  3. #3
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    Hello, soly_sol!

    1) Write the partial fraction decomposition of: . \frac{7x^2-x-12}{x(x+1)(x-1)}
    Why don't you try this method?


    We have: . \frac{7x^2-x-12}{x(x+1)(x-1)} \;=\;\frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}

    Then: . 7x^2 -x-12 \;=\;(x+1)(x-1)A + x(x-1)B + x(x+1)C


    Now substitute some "good" values for x.

    \begin{array}{ccccccc}\text{Let }x = 0: & -12 & = & (1)(\text{-}1)A + 0\!\cdot\!B + 0\!\cdot\!C & \Rightarrow &  A \:= \:12 \\<br />
\text{Let }x = \text{-}1: &-4 & = & 0\!\cdot\!A + (\text{-}1)(2)B + 0\!\cdot\!C & \Rightarrow & B \:= \:\text{-}2  \\<br /> <br />
\text{Let }x = 1: & -6 & = & 0\!\cdot\!A + 0\!\cdot\!B + 1\!\cdot\!2\!\cdot\!C & \Rightarrow & C \:= \:\text{-}3 \end{array}


    Therefore: . \frac{7x^2-x-12}{x(x+1)(x-1)} \;=\;\frac{12}{x} - \frac{2}{x+1} - \frac{3}{x-1}

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