About a statement of partial fraction in an answer

Question

I'm reading this answer of The logic behind partial fraction decomposition, I think my question is too basic and not directly related to the answer so I don't comment there. I don't understand why:

\begin{align} f(x) - \frac{c_r}{x - r} = \frac{1}{x - r} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right). \end{align} The expression in parentheses approaches $0$ as $x \to r$, and in fact it is a rational function whose numerator is divisible by $x - r$, so we can actually divide by $x - r$.

Why $\left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right)$ is divisible by $x-r$?

What I understand a little about the partial fraction is that it's because something forms a vector space so the $f(x)$ can be expressed/decomposed like that, but I'm not sure.

Answer 1

I'm not clear what the difficulty is. So, let's work through the algebra first.

Recall (first paragraph of the answer you cite), $f(x) = P(x)/Q(x)$ and $Q(x) = (x-r)R(x)$ and also (second paragraph) $c_r = P(r)/R(r)$. So, \begin{align*} f(x) - \frac{c_r}{x-r} &= \frac{P(x)}{Q(x)} - \frac{c_r}{x-r} \\ &= \frac{P(x)}{(x-r)R(x)} - \frac{c_r}{x-r} \\ &= \frac{P(x)}{(x-r)R(x)} - \frac{P(r)}{(x-r)R(r)} \\ &= \frac{1}{x-r} \cdot \frac{P(x)}{R(x)} - \frac{1}{x-r} \cdot \frac{P(r)}{R(r)} \\ &= \frac{1}{x-r} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right) \text{.} \end{align*}

So, perhaps the hiccup isn't the algebra, but is in why $f$, $Q$, and $c_r$ have the specified values?

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Your comments make me think we should cover why the facts I "recall" above are true.

• $f(x) = P(x)/Q(x)$: The Question introduced the fraction $P(x)/Q(x)$ and here we give it a name. It is given that $P$ and $Q$ are polynomials. So if $P$ is nonconstant, it has (complex) zeroes and likewise, non-constant $Q$ has zeroes.
• Let $n$ be the degree of $Q$ and $r_1, r_2, r_3, \dots, r_n$ be the roots of $Q$ (with repetitions, if $Q$ has any repeated root). Then it is a standard fact that $Q(x) = a(x-r_1)(x-r_2)(x-r_3)\cdots (x-r_n)$ for some constant $a$.
  You might be concerned that $Q$ has some other factorization that doesn't include $x-r$. But we know $Q$ is zero at $r$. The only way a product of things can be zero is when one (or more) of them are zero. So, any factorization of $Q$ includes a factor of $x-r$.
• We have already assumed that $P$ and $Q$ have no common factors. Why? If $P$ and $Q$ have common factors, we can "cancel" them, in matched pairs, one from $P$ and one from $Q$, until $P$ and $Q$ have no common factors. The effect of those common factors is to poke a hole in our graph.
  As an example, if $(x-s)^3(x-t)$ is a common factor of $P$ and $Q$, then we work with $P'(x)$ and $Q'(x)$ where $P'(x) = \frac{P(x)}{(x-s)^3(x-t)}$ and $Q'(x) = \frac{Q(x)}{(x-s)^3(x-t)}$, so $\frac{P(x)}{Q(x)} = \frac{(x-s)^3(x-t)P'(x)}{(x-s)^3(x-t)Q'(x)} = \frac{P'(x)}{Q'(x)}$ is an identity.
  (Note that being an identity means that the expressions all agree wherever they are all defined. $P'$ and $Q'$ are undefined at $s$ and $t$. In fact, $f = \frac{P}{Q}$ is undefined at the zeroes of $Q$ because division by zero is undefined. When we factor the common factors of $P$ and $Q$ out, we get a new, simpler expression, but the zeroes of $Q$ are still not in the domain of $f$, so they are "holes" in the graph of $f$. The coordinates of these holes are found by evaluating the cancelled fraction at the $x$-coordiante of the hole, $\left(s, \frac{P'(s)}{Q'(s)}\right)$ and $\left(t, \frac{P'(t)}{Q'(t)}\right)$. When we graph $P'(x)/Q'(x)$ it goes through these points. The graph of $P(x)/Q(x)$ is the same, except we delete those two points.)
• Having said all the above about cancelling common factors, it should be clear that $\left(r, \frac{P(r)}{Q(r)} \right)$ are the coordinates of a hole in the graph of $f$, so setting $c_r = \frac{P(r)}{Q(r)}$ is the right thing to do.