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Thread: long division using the remainder theorem

  1. #1
    Newbie
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    Red face long division using the remainder theorem

    we have just learnt this topic but i dont understand how i would do this:

    2x^4 + 3x^3 2x^2 + 4x 6

    divided by

    x^2 + x 2

    using the remainder theorem
    (aka in the form (Ax^2 + Bx + C)(x^2 + x 2) + Dx + E

    please help asap I have no idea how to do it, because there is a Dx and E at the end, and the power of 4 in the equn!

    thanksss
    xx
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by emydawn
    we have just learnt this topic but i dont understand how i would do this:

    2x^4 + 3x^3 2x^2 + 4x 6

    divided by

    x^2 + x 2

    using the remainder theorem
    (aka in the form (Ax^2 + Bx + C)(x^2 + x 2) + Dx + E

    please help asap I have no idea how to do it, because there is a Dx and E at the end, and the power of 4 in the equn!

    thanksss
    xx
    $\displaystyle 2x^4+3x^3-2x^2+4x-6\ =$ $\displaystyle (Ax^2+Bx+C)(x^2+x-2)+Dx+E$

    Now we may factor $\displaystyle x^2+x-2$ as follows:

    $\displaystyle x^2\ +\ x\ -\ 2\ =\ (x-1)(x+2)$

    Then when $\displaystyle x=1,$
    $\displaystyle 2x^4 + 3x^3-2x^2 + 4x-6\ =\ D\ +\ E$

    (can you see why?)

    and when $\displaystyle x=-2,$
    $\displaystyle 2x^4 + 3x^3-2x^2 + 4x-6\ =\ -2D\ +\ E$

    This gives a pair of simultaneous equations in $\displaystyle D$ and $\displaystyle E$, which I will leave for you to solve.

    RonL
    Last edited by CaptainBlack; Nov 22nd 2005 at 12:37 PM.
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