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    Hi could someone tell me how to divide

    x^5 -3x -1 by 5x^4 -3?
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    (Original post by Thaiprawns)
    Hi could someone tell me how to divide

    x^5 -3x -1 by 5x^4 -3?
    You can do this by long division.

    Make sure you write in the missing terms so 0x^4, 0x^3, 0x^2

    Take a read here: Click to read the article on division
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    Since you're dividing a polynomial of order 5 by a polynomial of order 4, you're going to get a resulting polynomial of order 1 (=5-4) and a remainder whose numerator is one less than the order of the polynomial in the denominator, i.e. 3 (=4-1). Algebraically, this means that
    \newline \dfrac{x^5-3x-1}{5x^4-3} \equiv Ax+B+\dfrac{Rx^3+Sx^2+Tx+U}{5x^4-3}

    If you multiply both sides by 5x^4-3, then you get
    \newline x^5-3x-1 \equiv (Ax+B)(5x^4-3) + Rx^3+Sx^2+Tx+U

    So you can expand the brackets and compare the coefficients to find the values of A,B,R,S,T,U that satisfy the above equation.

    The alternative way to do it (which might even be easier in this case) is polynomial long division, as gottastudy mentioned above.
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    I was using the long division way and for some reason find the first step hard! Would you guys mind giving me the first line please. I am fine when you're doing simple long division of polynomials. But would really like a hand with this type of one. Thanks guys so far
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    How many times does 5x^4 go into x^5?

    Normally the answer would be a nice multiple of x, but you get a fraction of x this time.
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    1/5(x)
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    Yep; so you say (5x^4 -3) goes into x^5 - 3x - 1 (1/5)x times.

    Multiply 5x^4 - 3 by (1/5) x and take that from the x^5 line, remembering as Gottastudy reminds you that the missing terms in x^4 etc are really there with 0 as the coefficient.
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    (Original post by Thaiprawns)
    1/5(x)


    Does this help?
 
 
 
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