Let $P$ be a polynomial of degree $n>1$, having n different real roots $x_{1},\dots ,x_n$ and a coefficient $a_n$ with $x^n$ and let $Q$ be a polynomial of degree not greater than $n-1$. Prove that $$ \frac{Q(x)}{P(x)} = \sum_{k=1}^{n} \frac{Q(x_k)}{P'(x_k)(x-x_k)}$$ for $ x \in R$, $x\notin$ {$x_1, \dots, x_n$}.
I have proved that $$P'(x_k)=a_n \prod_{j\not=k}(x_j - x_k)$$ but with that, I haven't got anything.
Since $P(x)$ has the form $P(x)=A(x-x_1)\dots (x-x_n)$, then $\dfrac{Q(x)}{P(x)}$ can be expressed in partial fractions whose existence can be seen in here, for example. So we have the following form $\displaystyle\frac{Q(x)}{P(x)} = \sum_{k=1}^{n} \frac{a_k}{(x-x_k)}$, then we multiply both members by $(x-x_j)$, $$ \frac{Q(x)}{P(x)} \cdot (x-x_j)= \sum_{k\neq j} \frac{a_k}{(x-x_k)}\cdot (x-x_j)+a_j $$ Now we do $x\to x_j$ and get $$ \lim_{x\to x_j}\frac{Q(x)}{P(x)} \cdot (x-x_j)=a_j $$ Hence $$ a_j=\lim_{x\to x_j}\frac{Q(x)}{P(x)} \cdot (x-x_j)=\lim_{x\to x_j}\left(\frac{x-x_j}{P(x)-P(x_j)}\right)Q(x)=\frac{Q(x_j)}{P'(x_j)},\quad \forall j=1,2,\dots,n. $$
Which concludes that: $$ \frac{Q(x)}{P(x)} = \sum_{k=1}^{n} \frac{a_k}{(x-x_k)} = \sum_{k=1}^{n} \frac{Q(x_k)}{P'(x_k)(x-x_k)} $$
