The Khan Academy proof of Polynomial Remainder Theorem (PRT) uses polynomial/algebraic long division:

$$ \begin{align} \frac{f(x)}{b(x)} =& q(x) &+ \frac{r(x)}{b(x)} \\ f(x) =& q(x)b(x) &+ r(x) \end{align} $$

If $b(x) = (x-a)$, then $f(a) = q(a)(a-a) + r(a) = q(a)\cdot 0 + r(a)$, and therefore $f(a) = r(a)$. Since $r(x)$ must be a constant (being lower degree than $(x-a)$), it's the same for all $x$ we get $f(a) = r$.

Isn't the first line undefined for division by $b(x)=x-a$ when $x=a$, because $b(a)=0$? i.e. it's divsion by zero?

  1. Is it OK, because in getting from the first line to the second line, it isn't "multiplied" by $b(x)$, but is "rewritten"? That is, it's a syntactic operation, not a numerical one, therefore we don't have to worry about divsion by zero. I don't think so, because then we could use the same argument for differentiation in calculus.

  2. Is it because limits always exist for polynomials with a factor cancelled out? This would mean we could skip epison-delta limits for polynomial differentiation (as I believe Newton did), and just say for $f(x)=x^2$, that $f'(x) = \frac{x^2+2ax+a^2 - x^2}{a} = 2x + a$ and as $a$ goes to zero, $f'(x)=2x$. Limits wouldn't be mentioned for PRT, because it's covered before limits have been introduced (in calculus).

If the second point is correct, PRT would (technically) be written with limits

$$\lim_{x\to a} f(x) = r$$


You just start from the second line.

Polynomial division is a process of decomposing a polynomial $f(x)$ into a sum $q(x)b(x)+r(x)=f(x)$. As you can see, you don't divide anything here in an arithmetical sense. All you do is performing an algorithm that produces the decomposition $q(x)b(x)+r(x)$ out of $f(x)$, the only worry you could have is that you divide by zero in the algorithm; but that isn't the case.

Well, you could say that this is the 1. answer. I think that it is important to say that no arithmetical division is taking place at all, other than what's done in the algorithm. So you don't even write $\dfrac{f(x)}{b(x)}$.

You don't have to use limits, because you don't divide by anything. But you can if you feel comfortable with it. Polynomial division came earlier than limits in the history of mathematics, so you can view "not writing the first line" as a workaround to not using limits/divide by zero.

There is one more thing, which is more subtle. When using polynomial division, you don't divide numbers, you divide polynomials and polynomials aren't numbers. As long as you don't want to divide by the zero polynomial ($f(x)=0$) (because you can't divide by zero everywhere) then you are good to go. Polymonials are entities of their own, independent of "plugging in values". And your concern is that plugging in $a$ can make polynomial division meaningless (i.e. there would be a division by zero). But polynomials can divide just fine without any substitutions, and only after the process, when you are just left with the identity $f(x)=q(x)b(x)+r(x)$, you can plug in $a$.

In other words, you use a tool (the polynomial divison) from a polynomial world to do something in your "arithmetical" world. In the arithmetical world, "polynomials" are just functions that take a number and split out a number, and your concern is competely valid. But in a polynomial world, the polynomial $x-a$ isn't a zero polynomial, so it is competely safe to divide by it.

(The ending here sounds like a fairly tale; it was my attempt at describing polynomial rings with layman terms, hope it sounded at least plausible.)

  • $\begingroup$ Thanks. I think it will help when I fully grasp how polynomial division finds a factor - I can see that it always works (eliminating the present leading term just adjusts the others), and you could continue with negative exponents (if wanted)... but not how it gets a factor, except that it's division. [P]olynomials aren't numbers is like my note 1 - couldn't the same argumennt apply to calculus: we aren't doing arithmetic (yet), it's just polynomials, so we can cancel out the a (as in my note 2. I feel calculus would be easier to understand if taught that way. $\endgroup$ – hyperpallium May 6 '18 at 10:04
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    $\begingroup$ In calculus, you are doing arithmetic. The division in the definition of a derivative is a true arithmetic division. It is because you should be able to calculate the derivative of any function, not only polynomials. If you are dealing with polynomials only then... well, you are right. But it is rare. $\endgroup$ – mz71 May 6 '18 at 10:11

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