<s>[INST] Why $(x-2)^2$ cannot be divided by $(x-2)(x-3)$ I know the question may seem ridiculous. My question is:

Why can't $(x-2)^2$ be divided by $(x-2)(x-3)$?

I know the answer is obvious if we do the division by hand. However, we also know that if remainder of $\frac{P(x)}{Q(x)}$ is zero, then $Q(x)$ is a divisor of $P(x)$.
Notice that:
$$(x-2)^2 = (x-3)(x-2)Q(x) + R(x).$$ If x =2 then 
$$(2-2)^2 = (2-3)(2-2)Q(x) + R(x) \to R(x) = 0,$$ so $(x-2)^2$ can be divided by $(x-2)(x-3)$.
This happens because $R(x) = ax +b$, and it isn't a single number, like $b$. Is there is another reason for it?
Thanks.
 [/INST] If you divide $(x-2)^2$ with $(x-2)(x-3)$ you get $\frac{(x-2)}{(x-3)}$. The polynomial $q(x)$ divides the polynomial $p(x)$ if there  exist a polynomial $g(x)$ such that $p(x) = g(x)\times q(x)$. You have that $\frac{(x-2)}{(x-3)} = \frac{1}{x-3} + 1$.
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