Show that this polynomial is reducible I'm trying to prove this question:
Show that $p(x)=x^3 + ax^2 + bx +1 \in \mathbb Z [x]$ is reducible over $\mathbb Z$ if and only if either $a=b$ or $a+b=-2$.
I did the converse in this way: 
if we take $a=b$, we see easily that $p(-1)=0$, so $p(x)$ has a root over $\mathbb Z$, then $p(x)$ is reducible over $\mathbb Z$.
I'm having problems with the first implication, I need some hints.
 A: If $a+b = -2$, you can see that $p(1) = 0$. Now a polynomial of degree $3$ is reducible over $\mathbb Z$ if and only if it has a root. But you know by the rational root theorem that if $p(x)$ has a root over $\mathbb Q$, then that root, write it $q/r$, has to be such that $q \, | \, a_0$ and $r \, | \, a_3$, where $p(x) = a_3 x^3 + \dots + a_0$. It follows that the only two possibilities are $+1$ and $-1$. 
The conditions $a=b$ or $a+b = -2$ correspond to the cases $+1$ and $-1$. I would've given an hint but stating the rational root theorem is pretty much giving the answer, so I thought I might as well do it.
Hope that helps,
A: If $p$ is reducible then there are polynomials $q_1$, $q_2$ of lower degree such that $p=q_1q_2$. Now, since $p$ has degree 3, one of $q_1$,$q_2$ must have degree 2, and the other degree 1. Thus, if $p$ is reducible, there are constants such that $$
  p(x) = (a_2x^2 + a_1x + a_0)(b_1x + b_0)
$$
Since the coefficient of $x^3$ in $p$ is $1$, it must be that $a_2b_1 = 1$. And because the constant term of $p$ is $1$, you similarly get $a_0b_0 = 1$. Thus, if $p$ is reducible, it must have a zero at $1$ or $-1$. Which, in other words means $$
\begin{eqnarray}
  a + b + 2 &=& 0 &\text{ or} \\
  a - b &=& 0
\end{eqnarray}
$$
