Irreducibility of $x^4 + mx + p$ over $\mathbb Q$ If $m$ and $p$ are integers with $p$ prime and $p \neq 3$ show that $x^4+mx+p$ is irreducible in $\mathbb Q[x]$ if and only if has not root in $\mathbb Q$.
 A: If your polynomial $f(x)=x^4+mx+p$ is reducible, then $f(x)=g(x)h(x)$, where either both $g$ and $h$ have degree 2, or one of them has degree $1$. The latter happens if and only if $f$ has a root in $\Bbb Q$. Hence you are left to prove that there are no $g$ and $h$ of degree $2$ such that $f=gh$.
Note that we can choose $g,h$ monic, up to multiplication by elements of $\Bbb Q$. Then to have
$$
\begin{align*}
f(x)&=(x^2+ax+b)(x^2+cx+d)\\
&=x^4+(c+a)x^3+(d+b+ac)x^2+(ad+bc)x+bd
\end{align*}
$$
we need
$$
\begin{cases}
a+c=0\\
d+b+ac=0\\
ad+bc=m\\
bd=p
\end{cases}
$$
Suppose that $b=p$ and $d=1$. Then the first two equations give
$$
\begin{cases}
c=-a\\
p+1=a^2
\end{cases}
$$
and the second clearly has solutions for $p=3$. We still need to prove that it has no solutions for $p\neq3$. For this, note that
$$
p+1=a^2 \Rightarrow p=a^2-1=(a+1)(a-1)
$$
which if $p\neq3$ implies that $p$ isn't prime, a contradiction.
A: Hint $\ $ Using undetermined coefficients, show that if it factored as two quadratics then the constant term would have form prime $\rm\: p = n^2-1,\:$ hence $\rm\:p = 3,\:$ contra hypothesis.
Remark $\ $ The hypothesis $\rm\:p\ne 3\:$ is necessary since $\rm\ x^4 - 4x + 3 = (x-1)^2 (x^2+2x+3).$
