Prove that the polynomial $p(x) = x^4 - 5x^3 + 3x - 2$ is irreducible over $\Bbb Q[X]$ I have proved that $p$ has no rational roots. Then I suppose that it factor in two quadratic polynomials $x^2 + ax + b, x^2 + cx + d$ to get a contradiction, but the system is hard. I think that there is an easy method to proof it. 
 A: The polynomial is irreducible over $\Bbb F_{11}$ and hence also over $\Bbb Q$. This is a bit easier to show with the equations, but still some work. 
A: The equations amount to solving
$$\tag1 a+c=-5,\quad b+d+ac=0,\quad bc+ad=3,\quad bd=-2$$
in integers.
The last eqution amounts to that one of $b,d$ is $\pm1$ and the other is $\mp2$. 
Then $b+d=\pm1$, hence $ac=1$, $a,c\in\{-1,1\}$, contradicting $a+c=-5$.
Apparently we can generalize: All
$$ x^4+ux^3+vx-2$$
with $u,v\in\Bbb Z$ and $|u|>2$ are either irreducible in $\Bbb Q[x]$ or have a root in $\{-2,-1,1,2\}$.
A: You have checked that $p$ has no rational roots, and therefore no roots in $\Bbb Z$, the ring of integers.
You have correctly concluded that the only possible $\Bbb Z$-factorization is as the product of two quadratics.
However, reading $p$ modulo $2$, we have $p(x)=x(x^3+x^2+1)$ as a polynomial over the field $\Bbb F_2$ with two elements. The second factor is $\Bbb F_2$-irreducible. Uniqueness of factorization in $\Bbb F_2[x]$ says that the only $\Bbb F_2$-factorization of $p$ is a linear times an irreducible cubic.
We’re done, for:
If there were a $\Bbb Z$-factorization of $p$ as product of two quadratics, this would go over to a factorization over $\Bbb F_2$ as a product of two quadratics. But there is no such $\Bbb F_2$-factorization, irrespective of reducibility of the quadratic factors.
