A confusion about the remainder theorem

Question

today, we were taught the polynomial remainder theorem and to prove it, the following identity was written in the board by our teacher.

$$f(x) = g(x)(x-A) + R ----------->(1)$$

Here, f(x) and g(x) are polynomials with f(x) being the dividend and (x-A) being the divisor. R is a constant with real value. (Please excuse my English for I have studied mathematics in my native language)

However, since I have knowledge in simplifying fractions, I thought that the same could be written as below,

$$f(x) / (x-A) = g(x) + R / (x-A) -------->(2)$$

In this case, we can not define the identity at x=A. (Or divide the polynomial at x=A for that matter). However we get the correct results from the connection I have marked as (1) above. How is this possible?

To further elaborate my question with a count example, we can do the following,

$$(x^3 + 6x^2 + 11x + 6) / (x+4) = x^2+2x+3-6/(x+4) ------->(3)$$

We know that we can not divide the polynomial at x = -4 in the connection I have marked as (3). However, we can write it as, $$(x^3 + 6x^2 + 11x + 6) = (x^2+2x+3-6)(x+4) - 6 -------->(4)$$

and then find the remainder by substituting x = -4

Can someone kindly explain to me how come this is possible?

Answer 1

So you're actually getting at a slightly subtle concept here: there is a distinction between polynomials as abstract objects and polynomials as functions. I'll answer your question from both perspectives:

1. Polynomials as abstract objects. In this case, we're working in a "world" where polynomials are the only things that exist. In that case, we can't write $f(x)/(x - A)$, since that isn't a polynomial (unless it turns out that $f(x)$ is a multiple of $x - a$, in which case we can write it, and everything works out fine, it just turns out that $R = 0$). Since we're dealing with polynomials as abstract objects here, the "what about when $x = A$ problem simply never comes up, because $x$ isn't a variable taking values in the real numbers - it's an extra element that we're adding to the real numbers so that we can write our polynomials down. When we do polynomial division and the like, we're often implicitly taking this approach, even if we don't explicitly say so.
2. Polynomials as functions. In this case, we can write $f(x)/(x - A)$, unless $x = A$. Whenever $x$ is not $A$, that's just a number divided by a non-zero number, so it's an honest-to-goodness fraction. When $x = A$, however, that's division by zero, so it is not the case that these two things are equivalent. Rather, those two things are equivalent everywhere except at $x = A$, which is often good enough for what we're doing. And, indeed, in this case, we only actually need (1), not (2), so there's no issue.
   There is also another approach here: we can treat polynomials as functions taking values in either the one-point compactification of the reals. That sounds fancy, but really it just means "we can just stick in either some extra thing called $\infty$, and whenever our function would be undefined, we'll just go ahead and say it takes that new value. Then everything works out fine (for what we're doing here): we do indeed have $$\frac{f(x)}{x-A} = g(x) + \frac{R(x)}{x - A}$$ everywhere, it just so happens that at one point, those two things are both $\infty$.

Note that you can also translate between the two quite easily: we can do whatever work we can/need to with abstract polynomials, then shift over to treating them as functions when we need to evaluate them, and nothing goes wrong.