I'm reading Wiki's page on Netwon's Method for root finding. I understand more than one root can be found, however unsure how it is accomplished. For instance, if the polynomial is degree $n$ it will have $n$ roots. Does anyone know how to accomplish this?
$$ x_{0A} = a, x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$
I recall reading something about repeating the process from $x_{0B} = x - a / 2$, however unsure what the variables in this are.
Since this does provide one answer to your question (there are others), I'll make it official and walk you through an example. Suppose I want to find the roots of $f(x) = \sin x$ but somehow have no real idea how this function behaves. I calculate the recursion formula $$x_{n+1} = x_n - \frac{\sin x_n}{\cos x_n}=x_n - \tan x_n$$ I arbitrarily choose $x_0 = 1$ and start calculating $$\begin{array}{rl}x_0&\phantom{-}1\\x_1&-0.557407724654902\\x_2&\phantom{-}0.065936451924841\\x_3&-0.000095721919325\\x_4&\phantom{-}0.000000000000293\\x_5&\phantom{-}0\end{array}$$ (With perfect computing $x_5$ would not be zero. But in this special case, computing error has paid off with a direct hit.)
In general, after finding a root $r_0$, to get the next root, replace $f(x)$ with $f_1(x) = \frac{f(x)}{x-r_0}$. Then $$f_1'(x) = \frac{f'(x)(x - r_0) - f(x)}{(x - r_0)^2}$$ And the recursion formula becomes $$x_{n+1} = x_n - \frac{f_1(x_n)}{f_1'(x_n)} = x_n - \dfrac {(x_n - r_0)f(x_n)}{f'(x_n)(x_n - r_0) - f(x_n)}$$
For my example, $r_0 = 0$, so $f_1(x) = \frac {\sin x}x$. This isn't defined at $0$, but $\lim_{x \to 0} \frac{\sin x}x = 1$, so if we needed to, we could define $f_1(0) = 1$. But there really isn't a point. We've already found $0$ and are looking for a different root. Why would we want to put the root we've already found back in? It might turn up as a point in the recursion, but this is so unlikely I wouldn't bother with it.
So I calculate the new recursion formula: $$x_{n+1} = x_n - \frac {x_n\sin x_n}{x_n\cos x_n - \sin x_n} = x_n - \dfrac {1}{\cot x_n - \frac 1{x_n}}$$ And arbitrarily start off with $x_0 = 1$ again: $$\begin{array}{rl}x_0&1\\x_1&3.79401891249195 \\x_2&2.83731861081935\\x_3&3.12005206105673\\x_4&3.14144824538471\\x_5&3.14159264695285\\x_6&3.14159265358979 \end{array}$$ An interesting alternative approach is to just do one or two iterations of the new formula, then switch back to the original $x_{n+1} = x_n - \tan x_n$: $$\begin{array}{rl}x_0&1\\x_1&3.79401891249195 \\\hline x_2&3.14205845439180\\x_3&3.14159265355610\\x_4&3.14159265358979 \end{array}$$
Why does this work? Remember that $f_1$ is just $f$ modified to remove the first root. The roots of $f_1$ are also roots of $f$, and with the exception of that first root, the roots of $f$ are roots of $f_1$. The problem is that my starting value of $1$ happens to be in a region where the original formula converges to the root $0$. Different starting values may converge to other roots, but I have no idea what values to try. For well-behaved functions, Newton's method should move me towards a root. So using the 2nd formula for a single iteration moves me from $1$ to $\sim 3.794$, which is close enough to another root for the original iteration to converge to $\pi$ instead of $0$.