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Math Help - Long Division Polynomials

  1. #1
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    Long Division Polynomials

    (2x^3+5x^2-4x-3) / (2X +1) = 2X^3 + 5X^2-4X-3 Remainder zero

    The back of the book says the remainder is -2

    I checked over my work a bunch of times, and used google cant find anything help would be much appreciated
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  2. #2
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    Re: Long Division Polynomials

    Quote Originally Posted by waleedrabbani View Post
    (2x^3+5x^2-4x-3) / (2X +1) = 2X^3 + 5X^2-4X-3 Remainder zero

    The back of the book says the remainder is -2

    I checked over my work a bunch of times, and used google cant find anything help would be much appreciated
    The remainder theorem states that when a polynomial \displaystyle \begin{align*} P(x) \end{align*} is divided by \displaystyle \begin{align*} ax + b \end{align*}, then the remainder is equal to \displaystyle \begin{align*} P\left(-\frac{b}{a}\right) \end{align*}.

    So here \displaystyle \begin{align*} P(x) = 2x^3 + 5x^2 - 4x - 3 \end{align*} and you are dividing by \displaystyle \begin{align*} 2x + 1 \end{align*}, so the remainder will be equal to \displaystyle \begin{align*} P\left(-\frac{1}{2}\right) \end{align*}.

    \displaystyle \begin{align*} P\left(-\frac{1}{2}\right) &= 2\left(-\frac{1}{2}\right)^3 + 5\left(-\frac{1}{2}\right)^2 - 4\left(-\frac{1}{2}\right) - 3 \\ &= 2\left(-\frac{1}{8}\right) + 5\left(\frac{1}{4}\right) + 2 - 3 \\ &= -\frac{1}{4} + \frac{5}{4} - 1 \\ &= 0 \end{align*}

    I agree with your answer of 0, so either you have copied down the wrong question, or the book has a typo.
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