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Math Help - Polynomial long division.

  1. #1
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    Polynomial long division.

    Can someone help me to simplify this expression:
    (x^8+x^6+x^4+x^2+1)/(x^4+x^3+x^2+x+1).
    The solution is x^4-x^3+x^2-x+1, but I don't know how to get it.
    Last edited by mr fantastic; June 5th 2011 at 12:05 PM. Reason: Re-titled.
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  2. #2
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    Just use long division.
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  3. #3
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    In the instruction it is said that I should use this formula: 1+q^2+q^3+...+q^n=(1-q^n+1)/(1-q). And it is ok for denominator, but I don't know what I should to do with numerator.
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  4. #4
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    x^8+x^6+x^4+x^2+1 = (x^2)^4 + (x^2)^3 + (x^2)^2 + (x^2) + 1

    let q = x^2 and use your given formula
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  5. #5
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    Hello, Frannny!

    \text{Simplify: }\:\frac{x^8+x^6+x^4+x^2+1}{x^4+x^3+x^2+x+1}

    \text{The solution is: }\:x^4-x^3+x^2-x+1

    We have: . \frac{x^8+x^6+x^4+x^2+1}{x^4+x^3+x^2+x+1}


    \text{Multiply by }\frac{x^2-1}{x^2-1}\!:\;\;\frac{x^2-1}{x^2-1}\cdot \frac{x^8+x^6+x^4+x^2+1}{x^4+x^3+x^2+x+1}

    . . . . . . . . . . . . . =\;\frac{x^{10}-1}{(x^2-1)(x^4+x^3+x^2+x+1)}


    \text{Multiply by }\frac{x-1}{x-1}\!:\;\;\frac{x-1}{x-1}\cdot \frac{x^{10}-1}{(x^2-1)(x^4+x^3+x^2+x+1)}

    . . . . . . . . . . . . . =\;\frac{(x-1)(x^{10}-1)}{(x^2-1)(x^5-1)}


    \text{Factor: }\;\frac{(x-1)(x^5-1)(x^5+1)}{(x-1)(x+1)(x^5-1)}

    \text{Reduce: }\;\frac{x^5+1}{x+1}


    \text{Factor: }\;\frac{(x+1)(x^4-x^3+x^2-x+1)}{x+1}

    \text{Reduce: }\;x^4-x^3+x^2-x+1

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