I want to decompose the rational function
$$ \cfrac{P(s)}{Q(s)}=\cfrac{\prod\limits_{i=1}^m (s+a_i)}{\prod\limits_{i=1}^n (s+b_i)} $$
where $a_i>0$ for every $i=1,\dots,m$, $b_i>0$ for every $i=1,\dots,n$ and $n>m$.
In other words, I'm looking for coefficients $x_j$, $j=1\dots,n$, such that
$$ \cfrac{P(s)}{Q(s)}=\cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n}\quad\Longleftrightarrow\quad \sum_{j=1}^n x_j \prod_{\substack{i=1\\i\neq j}}^n (s+b_i) = \prod_{i=1}^m (s+a_i) $$
Making some attempts with Mathematica for low values of $m$ and $n$ it seems that the $x_j$ are given by
$$ x_j=\cfrac{\prod\limits_{i=1}^m (a_i-b_j)}{\prod\limits_{\substack{i=1\\i\neq j}}^n (b_i-b_j)} $$
So my questions are:
1) Can I say beforehand that there exist unique such $x_j$s?
2) How can I prove that $x_j$ has in general (as it seems) the form above?
These are real (or complex) polynomials? This is known as "partial fractions", perhaps first seen by the typical student in elementary calculus courses.
(1) Suposing that your $P(s)/Q(s)$ in lowest terms, the necessary and sufficient condition that there exists $x_1,\dots,x_n$ such that $$ \cfrac{P(s)}{Q(s)}=\cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n} $$ Is: $m<n$ and no repeats among $b_1, b_2, \dots, b_n$.
(2) If such $x_j$ exist, they are unique.
Let's show $x_1$ is unique. Suppose $$ \cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n} =\cfrac{u_1}{s+b_1}+\dots+\cfrac{u_n}{s+b_n}. $$ Multiply by $s+b_1$, then take the limit as $s \to -b_1$. Result: $x_1=u_1$. Uniqueness for the others is the same.
(3) They have the form you describe.
Let's prove it for $x_1$. We have $$ \cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n} = \cfrac{\prod\limits_{i=1}^m (s+a_i)}{\prod\limits_{i=1}^n (s+b_i)} $$ Multiply by $s+b_1$ to get $$ x_1+ (s+b_1)\left(\cfrac{x_2}{s+b_2}+\dots+\cfrac{x_n}{s+b_n}\right) = \cfrac{\prod\limits_{i=1}^m (s+a_i)}{\prod\limits_{i=2}^n (s+b_i)} $$ Take the limit as $s \to -b_1$ [remembering that $b_j \ne b_1$ for all other $b_j$ ] $$ x_1 + 0 = \cfrac{\prod\limits_{i=1}^m (-b_1+a_i)}{\prod\limits_{i=2}^n (-b_1+b_i)} $$ Does that match your claimed formula?