1. ## Newton's Method

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

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

As an initial guess, I used x=1

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

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

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

Can someone help me see how they got from here:

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

to here:

$\displaystyle \frac{2x^3+5}{3x^2+2}$

2. $\displaystyle \displaystyle x - \frac{x^3+2x-5}{3x^2+2} =\frac{x(3x^2+2)}{3x^2+2} - \frac{x^3+2x-5}{3x^2+2} =\frac{x(3x^2+2)-(x^3+2x-5)}{3x^2+2}= \frac{2x^3+5}{3x^2+2}$

Do you follow?

3. 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!