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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.

    One thing though. Doesn't that general equation not leave space for the B, which is a constant?
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    (Original post by Student1256)
    Now that's sexy
    Seriously I'm done with my A2 maths and even I don't get it.
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    (Original post by RDKGames)
    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.
    If you give this guy rep I deserve some too. Same answer but further up. Just cos he can use TeX
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    (Original post by Jane122)
    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
    Hahahahahahahahaha this is some good ass trolling my friend
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    (Original post by Jane122)
    Attachment 612472

    Finally someone intelligent !
    Right that's wrong so we need to go back a step.

    Can you factorise f(x) given the information provided in part a)?
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    (Original post by Student1256)
    One thing though. Doesn't that general equation not leave space for the B, which is a constant?
    It does, the constant is part of the function R_1(x) when referring to the theorem.

    Remember, a general polynomial is of form f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0 for a_i\in \mathbb{Q} where a_0 is a constant so it is included.
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    (Original post by Jane122)
    I swear nobody on this thread bothered to look AT THE PIC.
    If you notice your pic is sideways, which isn't necessarily your fault (TSR Is annoying about it) but there is a correlation between that and your post, I suppose.

    One thing you can do to guarantee correct orientation is to upload the image to Imgur and then copy the location and use img tags, or if that fails (it is a bit too long winded to be honest unless you have done it before) is to take multiple by changing the orientation of the phone/camera device and then find one that makes them right way up.
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    (Original post by Hamzah249)
    Seriously I'm done with my A2 maths and even I don't get it.
    Yeah it took me a while to get it but that's how they explain sh*t in undergraduate level math at some unis. It's quite a pain to understand sometimes but you get used to the format and wording after a while hamxa bhai. I'm in a level
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    (Original post by Student1256)
    One thing though. Doesn't that general equation not leave space for the B, which is a constant?
    I was thinking the same thing.
    And looking at examples it's kinda hit and miss whether the constant is included. (I understand it's only in (x+n)^a)
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    (Original post by SeanFM)
    If you notice your pic is sideways
    I have a meme for that.
    Attached Images
     
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    (Original post by Student1256)
    Yeah it took me a while to get it but that's how they explain sh*t in undergraduate level math at some unis. It's quite a pain to understand sometimes but you get used to the format and wording after a while hamxa bhai. I'm in a level
    Yeah I got it too after a while, just from a stand point of someone who can't get through partial fractions, it's basically impossible to get what he wrote.

    Oh are you taking any modules in this month?
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    (Original post by RDKGames)
    It does, the constant is part of the function R_1(x) when referring to the theorem.

    Remember, a general polynomial is of form f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0 for a_i\in \mathbb{Q} where a_0 is a constant so it is included.
    Ok so am i right in saying that this formula covers for the existence of a constant. But doesn't have any way of affirming when it is 0 (ie no constant). So in a question when we aren't given a format for the answer (like a/(x+g) + b/(x+h) + c), assume the existence of a constant?

    A partial fractions question, that is.
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    (Original post by RDKGames)
    It does, the constant is part of the function R_1(x) when referring to the theorem.

    Remember, a general polynomial is of form f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0 for a_i\in \mathbb{Q} where a_0 is a constant so it is included.
    I see! Pure mathematics is beautiful and good job with your work on TSR
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    (Original post by Hamzah249)
    Yeah I got it too after a while, just from a stand point of someone who can't get through partial fractions, it's basically impossible to get what he wrote.

    Oh are you taking any modules in this month?
    Hahahahahaha yeah for someone who is finding it difficult comparing coefficients (no offence to that person, I myself was pretty bad at maths) trying to understand that is like trying to decipher ancient hieroglyphs :laugh:
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    (Original post by carpetguy)
    Ok so am i right in saying that this formula covers for the existence of a constant. But doesn't have any way of affirming when it is 0 (ie no constant). So in a question when we aren't given a format for the answer (like a/(x+g) + b/(x+h) + c), assume the existence of a constant?

    A partial fractions question, that is.
    Yes it covers for the existence of a constant that is 0 or not.

    If you're not provided with a format, assume that there is a constant then you will eventually show through working out that it is either identically 0 or some other value.
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    (Original post by Mr M)
    Right that's wrong so we need to go back a step.

    Can you factorise f(x) given the information provided in part a)?

    I got 2x^2+3x+1 when I factorised it
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    (Original post by Student1256)
    Hahahahahahahahaha this is some good ass trolling my friend
    What are you on about ?
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    (Original post by Jane122)
    I



    I got 2x^2+3x+1 when I factorised it
    Right you need to factorise this quadratic too.

    You'll then get:

    \displaystyle \frac{10}{(2x+1)(x+1)(x+3)} = \frac{A}{2x+1} + \frac{B}{x+1} + \frac{C}{x+3}

    Now find A, B and C using your preferred method.
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    (Original post by Mr M)
    Right you need to factorise this quadratic too.

    You'll then get:

    \displaystyle \frac{10}{(2x+1)(x+1)(x+3)} = \frac{A}{2x+1} + \frac{B}{x+1} + \frac{C}{x+3}

    Now find A, B and C using your preferred method.
    But I thought you wouldn't write it in the form of 0/x+3 with a remainder of 2x^2+3x+1 and do partial fractions on on the 0/x+3
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    (Original post by Jane122)
    But I thought you wouldn't write it in the form of 0/x+3 with a remainder of 2x^2+3x+1 and do partial fractions on on the 0/x+3
    Nope. Try and get A, B and C but, if you really have no idea what is going on, you probably should ask your teacher to go through it with you again.
 
 
 
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