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Math Help - Newton's Method

  1. #1
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    Newton's Method

    Approximate the (real) roots of x^3+2x-5=0

    I used x_{n+1}=x - \frac{x^3+2x-5}{3x^2+2}

    As an initial guess, I used x=1

    x_{n+1}= 1 - \frac{1^3+2(1)-5}{3(1)^2+2} = 1-\frac{-2}{5}=\frac{7}{5}

    I then did the itenerations out to x_5 and got the correct answer. However, when looking at the book's answer they had this:

    x_{n+1}=x - \frac{x^3+2x-5}{3x^2+2} = \frac{2x^3+5}{3x^2+2}

    Can someone help me see how they got from here:

    x_{n+1}=x - \frac{x^3+2x-5}{3x^2+2}

    to here:

     <br />
\frac{2x^3+5}{3x^2+2}<br />
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  2. #2
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    \displaystyle x - \frac{x^3+2x-5}{3x^2+2} =\frac{x(3x^2+2)}{3x^2+2} - \frac{x^3+2x-5}{3x^2+2} =\frac{x(3x^2+2)-(x^3+2x-5)}{3x^2+2}=  \frac{2x^3+5}{3x^2+2}

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  3. #3
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    I do see it. For some reason I was thinking that they somehow factored something out. Stupid algebra mistakes are going to doom me if I keep making them. Thanks for the help!
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