A Universe of Sorts

Let $p(x) = a_0 + a_1 x + \dots + a_n x^n$
If $p$ divides all coefficients except for the highest one ( $a_n$ ), $a_0$ is $p$ -squarefree ( $p^2$ does not divide $a_0$ ), then $p(x)$ is irreducible.
That is, $p | a0, p | a_1$ , upto $p | a_{n-1}$ , $p \not | a_n$ , and finally $p^2 \not | a_0$ .
Then we must show that $p(x)$ is irreducible.
Suppose for contradiction that $p(x) = q(x)r(x)$ where $q(x) = (b_0 + b_1 x+ \dots + b_k x^k)$ and $r(x) = (c_0 + c_1 x + \dots c_l x^l)$ (such that $k + l \geq n$ , and $k > 0, l > 0$ ).
See that $a_0 = b_0 c_0$ . Since $p | a_0$ , $p$ must divide one of $b_0, c_0$ . Since $p^2$ does not divide $a_0$ , $p$ cannot divide both $b_0, c_0$ . WLOG, suppose $p$ divides $b_0$ , and $p$ does not divide $c_0$ .
Also see that since $a_n = (\sum_{i + j = n} b_i c_j)$ , $p$ does not divide this coefficient $\sum_{i + j = n} b_i c_j$ . Thus, at least one term in $\sum_{i + j = n} b_i c_j$ is not divisible by $p$ .
Now, we know that $p$ divides $b_0$ , $p$ does not divide $c_0$ . We will use this as a "domino" to show that $p$ divides $b_1$ , $b_2$ , and so on, all the way upto $b_k$ . But this will imply that the final term $a_n$ will also be divisible by $p$ , leading to contradiction.
To show the domino effect, start with the coefficient of $x$ , which is $a_1 = b_0 c_1 + b_1 c_0$ . Since $a_1$ is divisible by $p$ , $b_0$ is divisible by $p$ , and $c_0$ is not divisible by $p$ , the whole equation reduces to $b_1 c_0 \equiv_p 0$ , or $b_1 \equiv_p 0$ [since $c_0$ is a unit modulo $p$ ].
Thus, we have now "domino"'d to show that $p$ divides both $b_0, b_1$ .
For induction, suppose $p$ divides everything $b_0, b_1, \dots, b_r$ . We must show that $p$ divides $b_{r+1}$ .
Consider the coefficient of the term $xri$ , ie $a_r$ . This is divisible by $p$ , and we have that $a_r = b_0 c_r + b_1 c_{r-1} + \dots + b_r c_0$ . Modulo $p$ , the left hand side vanishes (as $a_r$ is divisible by $p$ ), and every term $b_0, b_1, \dots, b_{r-1}$ vanishes, leaving behind $0 \equiv_p b_r c_0$ . Since $c_0$ is a unit, we get $b_r \equiv_p 0$ .
Thus, every term $\{ b_i \}$ is divisible by $p$ , implying $a_n$ is divisible by $p$ , leading to contradiction.
Again, the key idea: (1) $b_0$ is divisible by $p$ while $c_0$ is not. (This uses $p | a_0$ and $p^2 \not | a_0$ ). (2) This allows us to "domino" and show that all $b_i$ are divisible by $p$ (This uses $p | a_i$ ). (3) This show that $a_n$ is divisible by $p$ , a contradiction. (This uses $p \not | a_n$ ).

§ Eisenstein Theorem for Checking Irreducibility