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    \dfrac{3x^3-2x^2-3x+6}{(x-1)^2(x+1)} = 3-\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{(x+1)}

    3x^3-2x^2-3x+6=3(x-1)^2(x+1)-A(x-1)(x+1)+B(x+1)+C(x-1)^2

    when x=1 B=2

    when x=-1 C =1

    When Comparing Coefficients of x^2 i obtained A=-6
    Which meant for my partial fractions i got:

    3-\dfrac{6}{x-1}+ \dfrac{2}{(x-1)^2}+\dfrac{1}{x+1}

    however the book totally ignored my value for A and got

    3+\dfrac{2}{(x-1)^2}+\dfrac{1}{x+1}

    Am i correct or is the book correct?
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    (Original post by joyoustele)
    \dfrac{3x^3-2x^2-3x+6}{(x-1)^2(x+1)} = 3-\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{(x+1)}

    3x^3-2x^2-3x+6=3(x-1)^2(x+1)-A(x-1)(x+1)+B(x+1)+C(x-1)^2

    when x=1 B=2

    when x=-1 C =1

    When Comparing Coefficients of x^2 i obtained A=-6
    Which meant for my partial fractions i got:

    3-\dfrac{6}{x-1}+ \dfrac{2}{(x-1)^2}+\dfrac{1}{x+1}

    however the book totally ignored my value for A and got

    3+\dfrac{2}{(x-1)^2}+\dfrac{1}{x+1}

    Am i correct or is the book correct?
    I got the same as the book when subbing in x values
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    (Original post by bruh2132)
    I got the same as the book when subbing in x values
    really? Okay i will have to see where i went wrong
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    (Original post by joyoustele)
    ...
    Might wanna check how you got that A again. It's not right.
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    check by putting a number for x into the original function and your new version. they should match.
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    (Original post by RDKGames)
    Might wanna check how you got that A again. It's not right.
    comparing coefficient of x^2

    -2=3-A+C

    C=1

    A=-2-4

    A=-6 ?? Is that not correct?
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    (Original post by joyoustele)
    comparing coefficient of x^2

    -2=3-A+C
    C=1
    A=-2-4
    A=-6
    That 3 should be negative... check it.
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    (Original post by RDKGames)
    That 3 should be negative... check it.
    Damn. I wrote positive by mistake instead of a negative.
    TY RDKGames
 
 
 

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