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    (Original post by RDKGames)
    Ax+B+\frac{C}{x+1}
    Can you [email protected] my third attachment
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    (Original post by RDKGames)
    Ax+B+\frac{C}{x+1}
    How do you figure out which general partial fraction form you need for a fraction?
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    (Original post by Jane122)
    Im a female. So shut up.
    Aren't you a feisty one.

    I literally just told you to simplify it and you will get 4A+B=12 and/or use x=1, 2
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    (Original post by Hamzah249)
    Aren't you a feisty one.

    I literally just told you to simplify it and you will get 4A+B=12 and/or use x=1, 2

    Dont assume then. Jane is not a male name, it's an idiotic assumption. I've already got the correct answer. Thanks.
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    Name:  IMG_4894.jpg
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Size:  409.5 KB

    With part B, I have this so far


    Help Hamzah249
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    (Original post by h3rmit)
    How do you figure out which general partial fraction form you need for a fraction?
    That's what I was actually asking
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    (Original post by Jane122)
    Dont assume then. Jane is not a male name, it's an idiotic assumption. I've already got the correct answer. Thanks.
    Why are you so frustrated?
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    (Original post by student1256)
    why are you so frustrated?
    because nobody is looking at my attachment and im stuck
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    (Original post by h3rmit)
    How do you figure out which general partial fraction form you need for a fraction?
    For a general polynomial function f_n(x) of order n it is a theorem that \displaystyle \frac{f_n(x)}{g_m(x)}=R_{n-m}(x)+\frac{Q_{p}(x)}{g_m(x)} for n\geq m where R,Q are polynomial functions of their respective degrees and p<m.
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    (Original post by Jane122)
    because nobody is looking at my attachment and im stuck
    Are you still stuck? I thought you understand now with rdk's help
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    (Original post by Muttley79)
    Please don't post full solutions - the OP would have got there
    No he wouldn't have
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    (Original post by Jane122)
    Name:  IMG_4894.jpg
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    With part B, I have this so far


    Help Hamzah249
    You call him idiotic then go on to tag him asking for help, really?... lmao
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    (Original post by Student1256)
    Are you still stuck? I thought you understand now with rdk's help
    I don't I understand what he wrote.

    (Original post by RDKGames)
    For a general polynomial function f_n(x) of order n it is a theorem that \displaystyle \frac{f_n(x)}{g_m(x)}=R_{n-m}(x)+\frac{Q_{m-1}(x)}{g_m(x)} for n\geq m where R,Q are polynomial functions of their respective degrees.
    I'm confused
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    (Original post by RDKGames)
    For a general polynomial function f_n(x) of order n it is a theorem that \displaystyle \frac{f_n(x)}{g_m(x)}=R_{n-m}(x)+\frac{Q_{m-1}(x)}{g_m(x)} for n\geq m where R,Q are polynomial functions of their respective degrees.
    Now that's sexy
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    (Original post by Jane122)
    Name:  IMG_4894.jpg
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    With part B, I have this so far


    Help Hamzah249
    You've shown in part a that -3 is a root, meaning that (x-3) is a factor of f(x)=0.
    Divide f(x) by this using polynomial division (c2 stuff). Then you should have a nice quadratic to factorise. Following that it should be a simple partial fraction question without any of the nasty comparing coefficients needed.
    Edit: x+3 not x-3 pardon.
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    RDK already helped you with, gurl, you want 5 people to tell you same thing to get it through your head.
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    (Original post by Jane122)

    I'm confused
    That last one was not for you unless you fully understand the concept of partial fractions which doesn't seem like it at this point.

    In regards to your attachment, you have found the left over quadratic when you divided the cubic by x+3. Now factorise that quadratic then express your cubic in terms of 3 the linear factors you end up with.
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    (Original post by Student1256)
    You call him idiotic then go on to tag him asking for help, really?... lmao
    I didn't call him idiotic, I said his assumption was idiotic. There's a difference , if you weren't too busy flirting with him and RDK
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    (Original post by Jane122)
    Name:  IMG_4894.jpg
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    With part B, I have this so far


    Help Hamzah249
    Can you show your working please?
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    (Original post by Mr M)
    Can you show your working please?
    Finally someone intelligent !
 
 
 
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