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Math Help - Synthetic Division Help

  1. #1
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    Synthetic Division Help

    I usually know how to do Synthetic Division, but this one has my stumped. I'm not sure how to do it with the given divisor.


    Here's the equation

    (2x^3-17z^2+20x+6)/(2x-5)
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  2. #2
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    Quote Originally Posted by tony_rose88 View Post
    I usually know how to do Synthetic Division, but this one has my stumped. I'm not sure how to do it with the given divisor.


    Here's the equation

    (2x^3-17z^2+20x+6)/(2x-5)
    Synthetic division can only be done when the coefficient of x is 1. So:
    \frac{2x^3 - 17x^2 + 20x + 6}{2x - 5}

    = \frac{\frac{2x^3 - 17x^2 + 20x + 6}{2}}{\frac{2x - 5}{2}}

    = \frac{x^3 - \frac{17}{2}x^2 + 10x + 3}{x - \frac{5}{2}}

    So divide x^3 - \frac{17}{2}x^2 + 10x + 3 by <br />
x - \frac{5}{2}.

    -Dan
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  3. #3
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    Hello, Tony!

    You should have been shown how to handle these . . .


    (2x^3-17z^2+20x+6) \div (2x-5)
    The divisor must have a leading coefficient of 1.
    . . If the leading coefficient isn't 1, divide through by it.

    So we have: . 2x - 5 \quad\Rightarrow\quad x - \frac{5}{2}

    We will use \frac{5}{2} for our divisor.

    Important!
    Since we are dividing by one-half of the given divisor,
    . . the quotient will be twice as large as it should be.
    We will make the adjustment at the end.


    The synthetic division looks like this:

    . . . \begin{array}{cccccc}<br />
\frac{5}{2} & | & 2 & \text{-}17 & 20 & 6 \\<br />
& | & & 5 & \text{-}30 & \text{-}25 \\<br />
& & -- & -- & -- & -- \\<br />
& & 2 & \text{-}12 & \text{-}10 & \text{-}19 \end{array}


    The bottom says: . 2x^2 - 12x - 10,\quad \text{rem -}19

    But the quotient is twice as large as it should be ... remember?
    . .
    The remainder is unaffected.

    Therefore, the answer is: . x^2 - 6x - 5 -\frac{19}{2x-5}

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