Q:

Solve $f(x) = 6x^3 + 27x^2 + 17x + 20 \equiv 0 \pmod{30}$

Problem

Solve $f(x) = 6x^3 + 27x^2 + 17x + 20 \equiv 0 \pmod{30}$

My attempt was:
Since $30 = 2.3.5$, we then have:
$$
\begin{cases}
f(x) \equiv 0 \pmod{2}\\
f(x) \equiv 0 \pmod{3}\\
f(x) \equiv 0 \pmod{5}\\
\end{cases}
$$
By inspection, we see that:
$$
\begin{cases}
x \equiv 0 \pmod{2}\\
x \equiv 1 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 0 \pmod{5}\\
x \equiv 1 \pmod{5}\\
\end{cases}
$$
Hence, there will be four cases:
Case 1
$$
\begin{cases}
x \equiv 0 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 1 \pmod{5}\\
\end{cases}
$$
Case 2
$$
\begin{cases}
x \equiv 0 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 0 \pmod{5}\\
\end{cases}
$$
Case 3 
$$
\begin{cases}
x \equiv 1 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 1 \pmod{5}\\
\end{cases}
$$
Case 4
$$
\begin{cases}
x \equiv 1 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 0 \pmod{5}\\
\end{cases}
$$
Apply Chinese Remainder Theorem for the four system of equation, where $M = 30$:
$$
\begin{cases}
M_1 = \frac{30}{2} = 15\\
M_2 = \frac{30}{3} = 10\\
M_3 = \frac{30}{5} = 6\\
\end{cases}
$$
And, 
$$
\begin{cases}
15y_1 \equiv 1 \pmod{2} \implies y_1 = 1\\ 
10y_2 \equiv 1 \pmod{3} \implies y_2 = 1\\ 
6y_3 \equiv 1 \pmod{5} \implies y_3 = 1\\ 
\end{cases}
$$
Therefore, the four solutions are:
$$
\begin{cases}
x_1 = 1.15.0 + 1.10.2 + 1.6.1 = 26\\
x_2 = 1.15.0 + 1.10.2 + 1.6.0 = 20\\
x_3 = 1.15.1 + 1.10.2 + 1.6.1 = 41\\
x_4 = 1.15.1 + 1.10.2 + 1.6.0 = 35\\
\end{cases}
$$
Edit
Add missing cases suggested by yunone
Case 5 
$$
\begin{cases}
x \equiv 0 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 2 \pmod{5}\\
\end{cases}
$$
Case 6
$$
\begin{cases}
x \equiv 1 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 2 \pmod{5}\\
\end{cases}
$$
And the last two solutions are:
$$
\begin{cases}
x_5 = 1.15.0 + 1.10.2 + 1.6.2 = 32 \equiv 2 \pmod{30}\\
x_6 = 1.15.1 + 1.10.2 + 1.6.2 = 47 \equiv 17 \pmod{30}\\
\end{cases}
$$
Am I in the right track? Any idea?  
Thanks,

A:

So far this looks like you're on the right track. You may want to reduce the solutions $41$ and $35$ modulo $30$ to $11$ and $5$, respectively. Also, note that $x\equiv 2$ is a solution to $f(x)\equiv 0 \pmod{5}$, so you should address those two extra cases to find a total of $6$ solutions modulo $30$.
As a side note, you can use the $\LaTeX$ code \cdot to produce the multiplication symbol. For example, $30=2\cdot 3\cdot 5$ produces $30=2\cdot 3\cdot 5$.

