Consider the division of 73 by 5. Commonly, one could think of the result as 14 with a remainder of 3. In equations, however, this would seem to be expressed as
$$\frac{73}{5}=14+\frac{3}{5}\tag{1}$$
This might be a question of semantics, but what is the remainder, $\frac{3}{5}$ or $3$?
It seems to me that the more intuitive way to express the division, which reflects the original common statement is:
$$73=14\cdot 5+3\tag{2}$$ Which says that 73 is 14 multiplied by 5 with a remainder of 3.
$(1)$ and $(2)$ also differ because if we were to write them in a more general way we'd get
$$\text{dividend}=\text{quotient} \cdot \text{divisor}+\text{remainder}\tag{3}$$
$$\frac{\text{dividend}}{\text{divisor}}=\text{quotient}+\frac{\text{remainder}}{\text{divisor}}\tag{4}$$
$(3)$ and $(4)$ differ because $(4)$ isn't defined for a divisor of zero, but $(3)$ is.
I don't know the relevant theorem for real numbers, but for polynomials there is the following theorem:
(Spivak Calculus, Ch. 3, Problem 7a) For any polynomial function $f$, and any number $a$, there is a polynomial function $g$, and a number $b$, such that for all $x$ $$f(x)=(x-a)g(x)+b\tag{5}$$
This basically says that given a polynomial $f$ and a factor $x-a$, there is a polynomial $g$ such that if we multiply $g(x)$ by the factor $x-a$ and add a remainder $b$, we get $f$.
It would seem to be more intuitive to interpret the theorem as saying that
for any factor $x-a \neq 0$ we can always divide $f$ by $x-a$ and obtain quotient and a remainder.
ie $$\frac{f(x)}{x-a}=g(x)+\frac{b}{x-a} \tag{6}$$
This latter interpretation doesn't seem to be equivalent to the theorem, because the theorem is true for all $x$, including $x=a$.
Can the theorem have been written using $6$ instead of $5$?
