Solution: We recall that $p(x) \in \mathbb{Z} [x]$ is irreducible (reducible) in $\mathbb{Q} $ if and only if it is irreducible (reducible) in $\mathbb{Z} $. Since, the degree of the polynomial $p(x) = x^3 + 3x^2 -8$ is $3$, it is reducible if and only if it has a root in $\mathbb{Z} $. Let $\alpha \in \mathbb{Z} $ be a root of $p(x)$. This implies $\alpha ^3 + 3 \alpha ^2 - 8 = 0 $ and hence $\alpha $ must divide $8$. The possible roots of $p(x)$ are $\pm 1, \pm 2, \pm 4, \pm 8$. Now we will compute the value of $p$ at each of these integers. \begin{align*} p(-1) & = 6 \neq 0 \text{ and } (1) = -4 \neq 0 \\ p(-2) & = 59 \neq 0 \text{ and } (2) = -57 \neq 0 \\ p(-4) & = 501 \neq 0 \text{ and } (4) = -499 \neq 0 \\ p(-8) & = 4073 \neq 0 \text{ and } (8) = -4071 \neq 0 \\ \end{align*} Thus, the polynomial is irreducible over $\mathbb{Z} $ and hence $\mathbb{Q} $.