Here is the problem:
I believe you can use Gauss' Lemma whose proof says that if there are polynomials $g(x)$ and $h(x)$ in $\Bbb Q[x]$ with $\deg g(x) < \deg f(x)$ such that $f(x) = g(x)h(x)$, then there are polynomials $G(x)$ and $H(x)$ in $\Bbb Z[x]$ such that $\deg G(x) = \deg g(x), \deg H(x) = \deg h(x)$ and $f(x) = G(x)H(x). I'm just not positive how to do that exactly.
Thanks!
When you have a monic polynomial in $f(x) \in \mathbb{Z}[x]$, any root $\frac{p}{q} \in \mathbb{Q}$ of $f$ is also in $\mathbb{Z}$. This is because (assuming $\gcd(p,q)=1$),
\begin{eqnarray} 0 &=& f(\frac{p}{q}) \\ &=& \sum_{k=0}^{n} a_k(\frac{p}{q})^k \\ &=& \sum_{k=0}^{n} a_k\frac{p^k}{q^k} \\ &=& \sum_{k=0}^{n} a_kp^kq^{n-k} \\ \end{eqnarray}
where in the last line we multiplied the equation ($=0$) by $q^n$. In the last line, only the summand $p^n$ does not have the factor $q$, thus we see that $q | p^n$. Since $\gcd(p,q)=1$, it must be that $q=1.$ That is, the root is an integer.
Let the root be $m \in \mathbb{Z}$,
$$0=m^n + a_{n-1}m^{n-1} + ... +a_1m+a_0$$
and since $m$ divides LHS and the first $n-1$ summands, $m$ must also divide $a_0$. $\square$