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.

Let us have a polynomial $f(X)$. I claim that $X-a$ divides $f$ if and only if $a$ is a root of $f$.
From Euclidean division, we get that $f(X) = q(X)(X-a) + r(X)$, where $\deg r <\deg (X-a) = 1$ or $r = 0$. 
Assume that $X-a$ divides $f$. Then $r = 0$, and $f(a)=q(a)(a-a) = 0$. 
Now assume that $a$ is root of $f$. Then $0 = f(a) = q(a)(a-a)+r = r$, so $X-a$ divides $f$.
Now, $X-3$ does not divide $(X-2)^2$ since $(3-2)^2 \neq 0$, so $(X-3)(X-1)$ can't divide it either.
