$\mathbb Z_n[x]$, $x+(n-1)$ is a factor of $x^m+(n-1)$ for all $m$ and $n$ 
Prove that: in $\mathbb Z_3[x]$, $x+2$ is a factor of $x^m+2$. In $\mathbb Z_n[x]$, $x+(n-1)$ is a factor of $x^m+(n-1)$ for all $m$ and $n$.

You can use long division and get that $x^m+2=(x+2)(x^{m-1}+...+x+1)$. Does this prove that $x+2$ is then a factor or do I need to prove something more?
Similarly, using long division on $x^m+(n-1)=(n-1)(x^{m-1}+...+x+1)$ (I believe). 

The natural surjective homomorphism 
$\phi_n: \Bbb Z \to \Bbb Z_n \tag1$
induces a homomorphism, also surjective, and also denoted here by $\phi_n$,
$\phi_n: \Bbb Z[x] \to \Bbb Z_n[x]. \tag 2$
Now in $\Bbb Z[x]$, we have
$x^m - 1 = (x - 1)(x^{m -1} + x^{m - 2} + \ldots + 1);  \tag 3$
thus
$\phi_n(x^m - 1) = \phi_n((x - 1)(x^{m -1} + x^{m - 2} + \ldots + 1))$
$= \phi_n(x - 1) \phi_n(x^{m -1} + x^{m - 2} + \ldots + 1). \tag 4$
The image of $x^m - 1$ under $\phi_n$ is
$\phi_n(x^m - 1) = x^m - \phi_n(1) = x^m + (n -1) \in \Bbb Z_n[x], \tag 5$
and the image of $x - 1 \in \Bbb Z[x]$ is
$\phi_n(x - 1) = x - \phi_n(1) = x + (n - 1) \in \Bbb Z_n[x]; \tag 6$
by virtue of (5) and (6), (4) becomes
$x^m + (n - 1) = (x + (n - 1))\phi_n(x^{m -1} + x^{m - 2} + \ldots + 1) \in \Bbb Z_n[x], \tag 7$
which shows that
$x + (n - 1) \mid x^m + (n - 1) \tag 8$
in $\Bbb Z_n[x]$.  
Note that we don't need to explicitly present $\phi_n(x^{m -1} + x^{m - 2} + \ldots + 1)$ to complete this result, but it is nevertheless worth noting that
$\phi_n(x^{m -1} + x^{m - 2} + \ldots + 1) = x^{m - 1} + x^{m - 2} + \ldots + 1 \in \Bbb Z_n[x]. \tag 9$
The preceding argument, by virtue of (8), demonstrates that
$x + 2 \mid x^m + 2 \tag{10}$
in $\Bbb Z_3[x]$.  Long division is not really necessary once one has (3), which is easy to verify by direct multiplication:
$(x - 1)(x^{m -1} + x^{m - 2} + \ldots + 1)$
$= x(x^{m -1} + x^{m - 2} + \ldots + x + 1) - 1(x^{m -1} + x^{m - 2} + \ldots + x + 1)$
$= x^m + x^{m -1} + \ldots  + x - x^{m - 1} - \ldots - x -1 = x^m -1 \tag {11}$
(11) avoids the use of long division.
In the above we have made use of the fact that
$n - 1 \equiv -1 \; \mod n. \tag {12}$
