Show that $p(x)=x^2+9x+6$ is irreducible in $\mathbb Q[x]$ according to the following criteria:
a) Eisenstein's criterion
b) Gauss' theorem
c) Irreducibility on $\mathbb Z_p$, $p$ prime.
$\textbf a)$ By Eisenstein's criterion, if we choose $p=3$, then
$p \nmid 1$,
$p \mid 9\ \ $ and
$p^2 \nmid 6$.
And so is irreducible in $\mathbb Q$.
$\textbf b)$ The possible whole roots of $p(x)$ are $\pm1, \pm2, \pm3$ and $\pm6$, as none of them satisfies the equation $p(x) = 0$, so $P(x)$ is irreducible in $\mathbb Z$, and so by Gauss's theorem, it is irreducible in $\mathbb Q$
$\textbf c)$ If we choose $p=5$, then $p(x)$ can be written as
$$\overline{p}(x)=\overline{1}x^2+\overline{4}x+\overline{1} \in \mathbb Z_5[x]$$
How $\overline{p}(x)$ has no root in $\mathbb Z_5[x]$ and $5\nmid$ dominant coefficient, then it is irreducible in $\mathbb Z_5[x]$. And so is irreducible in $\mathbb Q$.
Are these ideas correct?