Why is $T^4+6T^2-8T-3\in\mathbb Q[T]$ is irreducible ?
Obviously Eisenstein doesn't work, but can we do a substitution and obtain a form such that the criterion is applicable ?
We didn't cover it in the lecture, but I've seen now there's a so called rational root theorem, If I didn't understand it wrong, does it state:
For any polynomials with integer coefficients and having a rational solution, say $x=p/q$ must $p$ be a factor of $a_0$(here $-3$) and $q$ a factor of leading coefficient, (here $1$), so in this case there is no rational root $\implies$ we cannot factor it as $(T^3+...)(T+...)$, it must be of the form $(T^2+aT+b)(T^2+cT+d)$ am I wrong ?
EDIT: After the hint of ajotatxe. I got the following relations;
$\begin{cases}bd=-3\\ad+bc=-8\\ac+b+d=6\\a+c=0\end{cases}$
Substituting $c=-a$ and $d=\frac{-3}b$ in the second equation and multiplying it by $b$, I have:
$ab^2-8b+3b=0$, I assume that $a$ is rational and consider $b$ as an indeterminate, thus;
$\displaystyle b_{1,2}=\frac{4\pm\sqrt{16-3a^2}}{a}$
In both cases $b$ is irrational, contradiction.
Can somebody confirm it ?
For this kind of problems you can suppose $a,b,c,d\in\mathbb Z$; see Gauss' Lemma. Then the system of equations becomes more tractable since from $bd=-3$ you get only few cases for $b$ and $c$.