Show that $p(x) = x^5+5x^4+10x^3+10x^2-x-2$ is irreducible over $\mathbb{Q}$ and has exactly $2$ nonreal roots.
$p(x)$ is irreducible:
Observe that we may express $p(x)$ as $$p(x) = x^5+5x^4+10x^3+10x^2+5x+1-6x-3 = (x+1)^5-6x-3.$$ Since $-1 \in \mathbb{Q},$ it follows that if $p(x-1)$ is irreducible in $\mathbb{Q}[x],$ then $p(x)$ is irreducible in $\mathbb{Q}[x]$ also. We check that $p(x-1) = x^5-6x+3$ is irreducible. Since $3 \mid -6, \hspace{1mm} 3 \mid 3, \hspace{1mm} 3 \nmid 1$ and $3^2 \nmid 3,$ it follows that $p(x-1)$ is irreducible by Eisenstein's Criterion. Hence $p(x)$ is also irreducible.
Graphing the polynomial shows that $p(x)$ crosses the real axis three times, so at most $p(x)$ has $2$ complex roots (how to show that there exactly $2$?) Also, what is another way to show polynomials in $\mathbb{Q}$ have a specific amount of roots in $\mathbb{C}$?
