I found this theorem that seems to outline a way of finding roots of a polynomial that have a multiplicity. Problem is, I can't understand it.
Perhaps you guys can help!
Practical method of finding the equal roots:
Let $f(x)=X_1X_2^2X_3^3X_4^4X_5^5\ldots X_m^m$, where$$\begin{align*} & X_1\equiv\text{product of all the factors like }(x-a)\\ & X_2^2\equiv\qquad\qquad\text{,,}\qquad\qquad\text{,,}\qquad\hspace{7mm}(x-a)^2\\ & X_3^3\equiv\qquad\qquad\text{,,}\qquad\qquad\text{,,}\qquad\hspace{7mm}(x-a)^3\end{align*}$$ $$\begin{align*} & \text{Find all the greatest common measure of }f(x)\text{ and }f'(x)=F_1(x)\text{ say}\\ & \qquad\qquad\qquad\text{,,}\qquad\qquad\qquad\text{,,}\qquad\hspace{3mm}\qquad F_1(x)\text{ and } F_1'(x)=F_2(x),\\ & \qquad\qquad\qquad\text{,,}\qquad\qquad\qquad\text{,,}\qquad\qquad\hspace{3mm}F_2(x)\text{ and }F_2'(x)=F_3(x),\\ & \ldots\qquad\qquad\qquad\ldots\qquad\qquad\qquad\ldots\qquad\qquad\qquad\ldots\qquad\qquad\ldots\end{align*}$$ Lastly, the greatest common measure of $F_{m-1}(x)$ and $F'_{m-1}(x)=F_m(x)=1$. Next, perform the divisions$$\begin{align*} & f(x)\div F_1(x)=\phi_1(x)\text{ say},\\ & F_1(x)\div F_2(x)=\phi_2(x),\\ & \qquad\ldots\qquad\ldots\qquad\ldots\\ & F_{m-1}(x)\div 1=\phi_m(x)\end{align*}$$ And finally$$\begin{align*} & \phi_1(x)\div\phi_2(x)=X_1\\ & \phi_2(x)\div\phi_3(x)=X_2\\ & \ldots\qquad\ldots\qquad\ldots\\ & \phi_{m-1}(x)\div\phi_m(x)=X_{m-1}\\ & F_{m-1}(x)=\phi_m(x)=X_m\end{align*}$$
I'm not sure what to do. I can't quite understand the method at all. I'm assuming that the polynomial given is of the form $\displaystyle f(x)=\sum_{k=0}^na_kx^k$ because then the roots of multiplicity would become trivial.
However, if that were the case, then what would be $X_i^i$?
The quoted passage shows how to get $X_1, X_2, \dots$ starting from a generic $f(x)$ polynomial.
Underlying idea is that $x=a$ is a root of multiplicity $m$ if $f(a)=f'(a)=\cdots=f^{(m-1)}(a)=0$.
For a minimalistic step by step example (as asked), let:
$$ f(x)=x^3-5 x^2+8x-4 $$
$$F_1(x) = \gcd(\,x^3-5 x^2+8x-4\,, \;3x^2-10 x+8\,) = x-2$$
$$F_2(x)=\gcd(\,x-2,\, 1 \,) = 1$$
Therefore $m=2$ since $F_2(x)=1$.
$$\phi_1(x)=f(x)\div F_1(x)=x^2-3x+2$$
$$\phi_2(x) = F_1(x)\div F_2(x) = x-2$$
$$X_1 = \phi_1(x)\div\phi_2(x) = x-1$$
$$X_2= \phi_2(x)=x-2$$
Then, since it was established earlier that $m=2\,$:
$$f(x)=X_1\,X_2^2=(x-1)(x-2)^2$$