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    (Original post by Mr M)
    I have a meme for that.
    yes Mr M !!!! is Mr M (jr) your son?
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    (Original post by Emz99)
    yes Mr M !!!! is Mr M (jr) your son?
    No he's just a wannabe. Isn't that the case Jr?
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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.
    What's the difference between order and degree in this context?

    Also, for \displaystyle R_{n-m}(x) do you keep going down until you get to a zero power variable, e.g for \frac{x^4}{x+1} is it Ax^3+Bx^2+C+\frac{D}{x+1}

    Also, PRSOM

    Also, also,1st tex attempt, pls no bully
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    You can open up the brackets and match the coefficients of each x term to find either A or B first and then substitute the known into a simpler equation involving co-efficients that are simple like of x squared to find the other.
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    (Original post by h3rmit)
    What's the difference between order and degree in this context?

    Also, for \displaystyle R_{n-m}(x) do you keep going down until you get to a zero power variable, e.g for \frac{x^4}{x+1} is it Ax^3+Bx^2+C+\frac{D}{x+1}

    Also, PRSOM

    Also, also,1st tex attempt, pls no bully
    I use those two terms interchangeably, I'm referring to the same thing.

    What you said is true but what you've shown is not quite there. \frac{x^4}{x+1}=Ax^3+Bx^2+Cx+D+ \frac{E}{x+1}

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    (Original post by RDKGames)
    I use those two terms interchangeably, I'm referring to the same thing.

    What you said is true but what you've shown is not quite there. \frac{x^4}{x+1}=Ax^3+Bx^2+Cx+D+ \frac{E}{x+1}

    Cheers
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    (Original post by Mr M)
    No he's just a wannabe. Isn't that the case Jr?
    man like Mr M
 
 
 
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