The root multiset arises naturally from the factorization of the polynomial
$$\rm f(x) = (x-r)^j \cdots (x-s)^k g(x)\ \to\ \\{\, j\cdot r,\:\ldots,\: k\cdot s\,\\}$$
where $\rm\:g(x)\:$ has no roots over the coefficient ring.
More precisely, define $\rm\ e_r(f(x)) := max\\{n\in\Bbb N\ :\ (x\\!−\\!r)^n\\!\mid f(x)\,\ in\,\ \Bbb R[x]\\}.\:$ Then the root multiset is $\rm\ \\{e_r(f)\cdot r\ :\ r\in roots(f)\\},\:$ where $\rm\:n\cdot r\:$ denotes an element $\rm\:r\:$ of multiplicity $\rm\:n\:$ in a multiset.
Note that, over a domain, these linear factors are _unique_ , being products of primes $\rm\,x-r.\:$ This uniqueness implies that the root multiset is well-defined. It also implies that any two (correct) root-finding algorithms will compute the same multiset of roots. The answer does not depend on what order the algorithm discovers the roots (as it generally does in _nonunique_ factorization domains).