# Why can a rational function be put in form $P(x)/Q(x)=\sum_{i=1}^n\frac{P(a_i)}{Q'(a_i)(x-a_i)}$?

In preparation for Lagrange's Interpolation Theorem, I'm curious to know why a rational function can be put in the following form.

Suppose $Q$ is a polynomial with distinct roots $a_1,\dots,a_n$, and let $P$ be a polynomial with $\deg(P)<n$. Then why does $$\frac{P(x)}{Q(x)}=\sum_{i=1}^n\frac{P(a_i)}{Q'(a_i)(x-a_i)}?$$ I tried expanding the right hand sum over a common denominator, but it got horribly messy quickly. I also tried induction, assuming $Q$ has just one distinct root, and thus $Q=C(x-a_1)^k$. But then $\deg(P)<1$, so $P$ is constant, say $P=A$. But then $$\frac{P(a_1)}{Q'(a_1)(x-a_1)}=\frac{A}{kC(a_1-a_1)^{k-1}(x-a_i)}$$ which seems problematic if $k>1$. What's the slicker way to conclude this equality? Thanks,

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You are not interpreting "$Q$ is a polynomial with distinct roots" in the intended way. It is intended to mean that the polynomial has degree say $n$, and all of its $n$ complex roots are different from each other. So $Q(x)=(x-1)^2$ is not covered. Indeed, the result is in general false for this $Q$, as you may remember from partial fractions. – André Nicolas Jan 24 '12 at 4:15
That clears it up, thanks. – Gemma Jan 24 '12 at 4:17

Multiply through by $Q(x)$ and note that both sides are polynomials of degree less than $n$ agreeing at the $n$ points $x=a_i$.
How can you be sure the right hand side has degree less than $n$? Each summand $\displaystyle\frac{P(a_i)Q(x)}{Q'(a_i)(x-a_1)}$ has degree $\deg(Q)-1$, but what if $\deg(Q)>n$? – Gemma Jan 24 '12 at 3:58