How to solve modular equivalence with polynomial I'm finding the solutions for the modular equation $x^2 -3x+2 \equiv 0\pmod{14}$. This is what I've done so far:
\begin{align}
0 & \equiv x^2 -3x+2 \pmod{14}\\
& \equiv (x-1)(x-2) \pmod{14} \\
\end{align}
This implies that two of the solutions are the polynomial's usual roots: 1, 2.
This is where I get lost. I realize I should be using the following fact to find the other solutions, but don't know how to approach it.$$(x-1)(x-2) \overset{14}{\equiv} 0 \Leftrightarrow 14|(x-1)(x-2)$$ The provided answer says that 8,9 are the other solutions. How would I go about finding these?

If $14|(x-1)(x-2)$, we cannot conclude $14|x-1$ or $14|x-2$, because $14$ is not prime.
But $14=2\times7$, so $14|(x-1)(x-2)$ means $2|(x-1)(x-2)$ and $7|(x-1)(x-2)$.  
Since $2$ and $7$ are prime, this means ($2|x-1 $ or $2|x-2$) and ($7|x-1$ or $7|x-2$). 
Thus, there are four solutions (mod $14$).  Can you take it from here?
