Originally Posted by

**peteryellow** Can somebody please check if the follwoing argument is correct:

Show that $\displaystyle x^4-4x-1$ is irrdeucible over rationals.

By the rational root theorem it has no rational roots.

Assume that

$\displaystyle x^4-4x-1 = (x^2+ax+b)(x^2+cx+d)$ then

$\displaystyle a+c=0,\ d+ac+b=0\ ad+bc=-4\ bd=-1.$

$\displaystyle a=-c$ implies that $\displaystyle d+b-c^2=0$,

$\displaystyle c(b-d)=-4$, $\displaystyle bd=-1$ Correct as far as here!

c=-4 or b-d=-4. NO!!!

Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.

c=-4 shows that d+b=16 which is a contradiction.

if b-d=-4 we also get a contradiction. So the polynomial is irreducible.

Is it correct?