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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 $X_i^i$?

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    Work it out step by step for $m=2$ and it will become more obvious. – 2017-01-12
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    @dxiv Okay, so assuming $f(x)=XX_2^2$, how would we take the derivative of that? Would we replace it with $(x-a)^2$ and use the chain rule? – 2017-01-12
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    No, it works the other way around. What they show you is how to get $X_1$ and $X_2$ starting from a generic $f(x)$ polynomial. – 2017-01-12
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    @dxiv I'm still confused... Can you just post an answer with an example? – 2017-01-12

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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:

$$ f(x)=x^3-5 x^2+8x-4 $$

Find all the greatest common measure of $f(x)\text{ and }f'(x)=F_1(x)$

$$F_1(x) = \gcd(\,x^3-5 x^2+8x-4\,, \;3x^2-10 x+8\,) = x-2$$

$\text{,,}\qquad F_1(x)\text{ and } F_1'(x)=F_2(x),$

$$F_2(x)=\gcd(\,x-2,\, 1 \,) = 1$$

Lastly, the greatest common measure of $F_{m-1}(x)$ and $F'_{m-1}(x)=F_m(x)=1$

Therefore $m=2$ since $F_2(x)=1$.

Next, perform the divisions $f(x)\div F_1(x)=\phi_1(x)$

$$\phi_1(x)=f(x)\div F_1(x)=x^2-3x+2$$

$F_1(x)\div F_2(x)=\phi_2(x)$

$$\phi_2(x) = F_1(x)\div F_2(x) = x-2$$

And finally $\phi_1(x)\div\phi_2(x)=X_1$

$$X_1 = \phi_1(x)\div\phi_2(x) = x-1$$

$F_{m-1}(x)=\phi_m(x)=X_m$

$$X_2= \phi_2(x)=x-2$$

Then, since it was established earlier that $m=2\,$:

$$f(x)=X_1\,X_2^2=(x-1)(x-2)^2$$

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    Wait, why does $\phi_3(x)=1$? – 2017-01-12
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    @Frank Sorry, I had quoted the wrong line from the question (edited and fixed now). There is no $\phi_3\,$, at the last step $\phi_m(x)=X_m$ i.e. for $m=2\,$: $\,\phi_2(x)=X_2\,$. – 2017-01-12
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    For some reason, this method doesn't work with perfect cubes, such as $f(x)=(x-1)^3$. When I go through the method, it gives $X_1=1,X_2=x-1$ – 2017-01-13
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    @Frank The method works, recheck your calculations. For $f(x)=(x-1)^3$ you get $F_1(x)=(x-1)^2\,$, $F_2(x)=(x-1)\,$, $F_3(x)=1\,$ and in the end $X_1=X_2=1\,$, $X_3=x-1$. – 2017-01-13