You could try the following: If $p$ is reducible, it will neccesarily reduce to a pair of linear factors. Furthermore, as $p$ is monic, one can assume that the factors are monic as well. Thus, if $p$ were reducible, $p = (x+a)(x+b)$. By comparison of coefficients, $a + b = 0$ and $ab = t$, so $b = -a$ and thus $-a^2=t$. Now, $a = q_1/q_2$, where $q_1, q_2$ are polynomials in $t$, so $-q_1^2/q_2^2 = t$, or equivalently, $-q_1^2 = t q_2^2$. But since $\deg q_1^2$ is even and $\deg t q_2^2$ is odd, no such factorization can exist. This is, essentially, a rigourous version of your argument.