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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.

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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?


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.
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  • $\begingroup$ But why the part in the parentheses is divisible by $x-r$? I'm not very good at math so may be apparent but I didn't see it... $\endgroup$ – user7813604 Dec 21 '18 at 1:54
  • $\begingroup$ @user7813604 The part inside the parentheses must be divisible by $x-r$ because we assumed that $f$ had a simple pole at $r$, and on the LHS we have removed it by subtracting off just that pole scaled appropriately by the value $c_r=P(r)/R(r)$ (notice the typo in this answer, and compare to the post you linked to in the OP). $\endgroup$ – DeficientMathDude Dec 21 '18 at 2:31
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    $\begingroup$ @user7813604 Alternatively, you can say that the part inside the parentheses must be divisible by $x-r$ because it is a rational function that goes to $0$ at $r$; if the numerator doesn't go to $0$ then the entire expression can't go to $0$, and the numerator going to $0$ by definition means that the numerator is divisible by the monomial $(x-r)$. $\endgroup$ – DeficientMathDude Dec 21 '18 at 2:35
  • $\begingroup$ @DeficientMathDude: Thanks! What I really want to know is your second comment, I wanted to ask about why the fact that "a rational function that goes to 0 at r" means "the numerator is divisible by the monomial (x-r)". Because I don't think in my current level I am able to understand the "simple pole at r" since I don't know about complex analysis (I want to avoid it). $\endgroup$ – user7813604 Dec 21 '18 at 3:30
  • $\begingroup$ @DeficientMathDude : Thanks for catching the typo. Fixed in all three places. $\endgroup$ – Eric Towers Dec 21 '18 at 18:06

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