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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    Hello,

    Could someone please help me by telling me what the best way to resolve the following equation into three partial fractions is?

    6x^2 +5x -2<br />
\div<br />
x(x-1)(2x+1)
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  2. #2
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    Quote Originally Posted by HNCMATHS View Post
    Hello,

    Could someone please help me by telling me what the best way to resolve the following equation into three partial fractions is?

    6x^2 +5x -2<br />
\div<br />
x(x-1)(2x+1)
    \frac{6x^2+5x-2)}{x(x-1)(2x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{2x-1}


    Which becomes 6x^2+5x-2 = A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1)

    It's best to find f(0), f(1) and f\left(-\frac{1}{2}\right)<br />
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  3. #3
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    Sorry I do not understand your reply
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  4. #4
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    \frac{A}{x}+\frac{B}{x-1}+\frac{C}{2x+1}=\frac{A(x-1)(2x+1)+Bx(2x+1)+Cx(x-1)}{x(x-1)(2x+1)}

    =\frac{6x^2+5x-2}{x(x-1)(2x+1)}

    Now, you can ignore the denominators and multiply out the numerator on the right and compare terms with the left to find A, B and C.

    The reason e^{i*pi} recommended now finding f(0), f(1) and f(-0.5) is because f(0) finds A since the parts with B and C will be zero.

    You get 6(0)+5(0)-2=A(-1)(1) and so on for B and C.
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  5. #5
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    For more examples of partial fractions:

    Partial fraction - Wikipedia, the free encyclopedia
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  6. #6
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    Helolo, HNCMATHS!

    I'll show you the method I was taught.
    I think it's much simpler; don't know why everyone doesn't teach it.


    Resolve into partial fractions: . \frac{6x^2 +5x -2}{x(x-1)(2x+1)}
    Set up the equation: . \frac{6x+5x-2}{x(x-1)(2x+1)} \;=\;\frac{A}{x} + \frac{B}{x-1} + \frac{C}{2x+1}


    Multiply by the LCD: . 6x^2 + 5x - 2 \;=\;(x-1)(2x+1)A + x(2x+1)B + x(x-1)C


    Select some "good" values for x\!:

    Let x=0: . . . . . 0 + 0 - 2 \:=\:(\text{-}1)(1)A + 0\!\cdot\! B + 0\!\cdot\! C . . . . . \Rightarrow . . . \text{-}A \:=\:\text{-}2\quad\Rightarrow\quad\;\,\boxed{ A \:=\:2}

    Let x = 1: . . . . 6+5-2 \:=\:0\cdot A + (1)(3)B + 0\cdot C . . . . . .  \Rightarrow\qquad 3B \:=\: 9 . . \Rightarrow\quad\;\boxed{ B\: =\: 3}

    Let x = \text{-}\tfrac{1}{2}:\;\;6\left(\tfrac{1}{4}\right) + 5\left(\text{-}\tfrac{1}{2}\right)-2 \:=\: 0\cdot A + 0\cdot B + \left(\text{-}\tfrac{1}{2}\right)\left(\text{-}\tfrac{3}{2}\right)C \quad\Rightarrow . . . \tfrac{3}{4}C \:=\:\text{-}3 \quad\;\;\Rightarrow\quad\boxed{ C \:=\:\text{-}4}


    Therefore: . \frac{6x^2+5x-2}{x(x-1)(2x+1)} \;=\;\frac{2}{x} + \frac{3}{x-1} - \frac{4}{2x+1}


    [Edit: oops! .I see that e^{i\pi} recommended this method.]

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  7. #7
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    Quote Originally Posted by HNCMATHS View Post
    Could someone please help me by telling me what the best way to resolve the following equation into three partial fractions is? 6x^2 +5x -2<br />
\div<br />
x(x-1)(2x+1)
    Look at this result. It was done with a CAS costing about 40$ US.
    There are even free programs, which will do this!
    The idea of doing partial fractions has gone the way of square roots and horseshoes.
    When has anyone done the square of 13 by hand?
    Attached Thumbnails Attached Thumbnails Partial Fractions-untitled.gif  
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    \frac{A}{x}+\frac{B}{x-1}+\frac{C}{2x+1}=\frac{A(x-1)(2x+1)+Bx(2x+1)+Cx(x-1)}{x(x-1)(2x+1)}

    =\frac{6x^2+5x-2}{x(x-1)(2x+1)}

    Now, you can ignore the denominators and multiply out the numerator on the right and compare terms with the left to find A, B and C.

    The reason e^{i*pi} recommended now finding f(0), f(1) and f(-0.5) is because f(0) finds A since the parts with B and C will be zero.

    You get 6(0)+5(0)-2=A(-1)(1) and so on for B and C.
    ***
    Hello Archie, I understand what you have done up until the line starting... Now , you can ignore the denominators ??

    Sorry

    Chris
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  9. #9
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    My main problem is that i dont understand how to work out that values that we need to plug in for x

    EDIT**

    I think I have worked it out, x has to equal 0 in the divisor? so in our case where we have 2x+1 ..... x would equal -0.5

    Its just clicked, again thank you all and i will be clicking thanks on all your posts.
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  10. #10
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    Quote Originally Posted by HNCMATHS View Post
    My main problem is that i dont understand how to work out that values that we need to plug in for x

    EDIT**

    I think I have worked it out, x has to equal 0 in the divisor? so in our case where we have 2x+1 ..... x would equal -0.5

    Its just clicked, again thank you all and i will be clicking thanks on all your posts.
    Glad to hear it
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