Polynomial of the nth degree and its elements. I have to prove, that polynomial:
$\ x^n + a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0=0$
, where $ a_i= [-1,0,1] $ and $ i=0,1,.....,n $
doesn't have any solutions in the set
$$
\begin{aligned}
( -\infty,-2)  \cup (2, \infty)
\end{aligned}
$$
We know, that $ a_n=1$, so we know, that $ x_1+x_2+....+x_n=-2$.
It's all I have though, any 
suggestions, what can I do in next step?

Let $f(x)=0$, then
\begin{align}
0=&|x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0|\\
\geq&|x|^n-|a_{n-1}||x|^{n-1}-\cdots-|a_1||x^1|-|a_0|
\end{align}
Write it as
\begin{align}
|x|^n\leq & |a_{n-1}||x|^{n-1}+\cdots+|a_1||x^1|+|a_0|\\
\leq & |x|^{n-1}+|x|^{n-2}+\cdots+|x|+1\\
= & \frac{|x|^{n}-1}{|x|-1}
\end{align}
If $|x|>2$, we get $|x|-1>1$ and then
$$
|x|^n\leq |x|^n-1.
$$
It is impossible.
