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Thread: Dividing polynomial

  1. #1
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    Dividing polynomial

    I'm having problems dividing the polynomial 2x^5 + 2x^3 + x^2 + 1 by x^2 + 1
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  2. #2
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    Looks straight forward to me. But I don't know what to tell you without seeing what you did. Did you remember to include "0x" in the divisor and "$\displaystyle 0x^4$" and "$\displaystyle 0x$" in the dividend? How many times does $\displaystyle x^2$ divide into $\displaystyle 2x^5$?
    Last edited by HallsofIvy; Mar 23rd 2011 at 04:49 AM.
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  3. #3
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    It goes in 2x^3 times but I get x^2 as the next quotient part but the correct quotient should be 2x^3 - 1 according to the answer the book gives
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  4. #4
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    $\displaystyle x^2+0x+1sqrt(2x^5+0x^4+2x^3+x^2+0x+1)$

    This is the subtracting parts: The quotient i got is $\displaystyle 2x^3+x^2+x+1$

    Part 1 $\displaystyle 2x^5+0x^4+2x^3$

    Part 2 $\displaystyle x^4+x^2$

    $\displaystyle x^4+x^3+x^2$

    Part 3 $\displaystyle x^3+0x$

    $\displaystyle x^3+x^2+x$

    Part 4 $\displaystyle x^2+1$

    $\displaystyle x^2+x+1$

    Final $\displaystyle x$
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  5. #5
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    Yes, $\displaystyle x^2$ divides into $\displaystyle 2x^5$ $\displaystyle 2x^3$ times. And multiplying the entire divisor by $\displaystyle 2x^3$ gives $\displaystyle 2x^5+ 2x^3$
    Now subtract that from the dividend:
    $\displaystyle 2x^5+ 2x^3+ x^2+ 1- (2x^5+ 2x^3)= x^2+ 1$.

    And now, of course, $\displaystyle x^2+ 1$ divides into that exactly once: the quotient is $\displaystyle 2x^3+ 1$.
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