I started like this
$$\begin{align}
120x + 120 = 0 &\rightarrow& x &= &-1\\
- x^{120} - x^{119} - ... - x^2 - x^1 +120 = 0 &\rightarrow&x &= &1\\
x + 120 = 0 &\rightarrow& x& = &-120\\
x^7 - x^3 + 120 = 0 &\rightarrow& x &= &-2\\
x - 2 = 0 &\rightarrow& x &= &2\\
\end{align}
$$
Now we have $-120, -2, -1, 0, 1, 2, 120$.
Having both $-1$ and $1$ means that I can always get the negated value of any new number.
But I can't crack this puzzle as I keep finding new numbers.
Wonder if I got it all wrong...
Edit:
Some more (thanks to Aravind for the first 3)...
$$
\begin{align}
x^2 + 2x - 120 = 0 &\rightarrow& x &= &10\\
2x - 10 = 0 &\rightarrow& x &= &5\\
10x - 120 = 0 &\rightarrow& x &= &12\\
x^4 + x^3 + x^2 + x - 120 = 0 &\rightarrow& x &= &3\\
\end{align}
$$
And
$$
\begin{align}
2x - 120 = 0 &\rightarrow& x &= &60\\
3x - 120 = 0 &\rightarrow& x &= &40\\
5x - 120 = 0 &\rightarrow& x &= &24\\
5x^4 + 10x - 120 &\rightarrow& x &= &4\\
4x - 120 = 0 &\rightarrow& x &= &30\\
4x^2 + 4x -120 &\rightarrow& x &= &5\\
5x -120 = 0 &\rightarrow& x &= &24\\
3x^2 + 2x - 120 = 0 &\rightarrow& x &= &6\\
6x - 120 = 0 &\rightarrow& x &= &20\\\\
-2x^2 - x + 120 = 0 &\rightarrow& x &= &-8\\\\
-8x + 120 = 0 &\rightarrow& x &= &15\\
\end{align}
$$
So far I have
$\{0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}$ (and all the negated numbers as well).
As far as I understand from other answers/comments this should be the complete list. I still wonder how that is proved.