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Thread: [SOLVED] long division/ Remainder Theorem

  1. #1
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    [SOLVED] long division/ Remainder Theorem

    help please

    Problem:

    Use long division and the Remainder Theorem to evaluate , where

    The quotient is
    The remainder is
    =
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  2. #2
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    Quote Originally Posted by lsnyder View Post
    help please

    Problem:

    Use long division and the Remainder Theorem to evaluate , where

    The quotient is
    The remainder is
    =
    Divide $\displaystyle x^3- 5x^2+ 8x- 6$ by $\displaystyle x-2$

    x divides into $\displaystyle x^3$ $\displaystyle x^2$ times and $\displaystyle x^3- 5x^2- x^2(x- 2)= x^3- x^3- 5x^2+ 2x^2= -3x^2$
    so we now have $\displaystyle -3x^2+ 8x- 6$ left.

    x divides into $\displaystyle -3x^2$ $\displaystyle -3x$ times and $\displaystyle -3x^2+ 8x- (-3x)(x-2)= -3x^+ 3x^2+ 8x- 6x= 2x$
    so we now have $\displaystyle 2x- 6$ left.

    x divides into $\displaystyle 2x- 6$ 2 times and $\displaystyle 2x- 6- 2(x-2)= 2x- 2x- 6+ 4= -2$

    The quotient is $\displaystyle x^2- 3x+ 2$ and the remainder is -2.

    (Check my arithmetic.)
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