(I'm actually learning calculus.)Before I started working on this problem,I went to read this proof: Partial Fractions Proof
I think I understand what the proof tried to do(And I can complete some proofs for simple cases like case 1 or case 2 by myself)
Then I changed the condition "If $r$ is a root of $g(x)$ with multiplicity 1" to "If $g(x)$ has n distinct roots which can be denoted as '$(x-r_{1} )(x-r_{2} )\dots (x-r_{n} )$' with multiplicity $1$",I started to prove: $$\frac{f(x)}{g(x)}=\sum_{i=1}^{n}\frac{A_{i} }{x-r_{i} }$$ $A_{i}$ is a constant,and $deg(f(x))<deg(g(x))$.
Then I tried to prove it like what the proof mentioned before did too.
Here is my attempt:
Since I need to prove:$$\frac{f(x)}{(x-r_{1})(x-r_{2})\dots(x-r_{n})}=\frac{A_{1}}{x-r_{1}}+\frac{A_{2}}{x-r_{2}}+\dots+\frac{A_{n}}{x-r_{n}}$$ I need to prove: $$f(x) = A_{1}(x-r_{2})(x-r_{3})\dots(x-r_{n})+A_{2}(x-r_{1})(x-r_{3})\dots(x-r_{n})+\dots+A_{n}(x-r_{1})(x-r_{2})\dots(x-r_{n-1})$$ (Equa1)
By Bezout's Identity I have: $$m_{1}(x)(x-r_{2})(x-r_{3})\dots(x-r_{n})+m_{2}(x)(x-r_{1})(x-r_{3})\dots(x-r_{n})+\dots+m_{n}(x)(x-r_{1})(x-r_{2})\dots(x-r_{n-1})=1$$ (Equa2)
Then multiply Equa2 with $f(x)$ I can get: $$\frac{f(x)}{g(x)}=\frac{m_{1}f(x)}{x-r_{1}}+\frac{m_{2}f(x)}{x-r_{2}}+\dots+\frac{m_{n}f(x)}{x-r_{n}}$$
Here is where I was stuck: by long division I could split every term into form like: $$q(x)+\frac{C_{i}}{x-r_{i}}$$
BUT IS THE QUOTIENT $0$? OR SHOULD I PROVE $m_{n}(x)f(x)$ is a constant?
