Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but that it has two different irreducible factors in $\mathbb{R}[X]$ 
Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$.

I've tried to use the cyclotomic polynomial as:
$$X^5-1=(X-1)(X^4+X^3+X^2+X+1)$$
So I have that my polynomial is 
$$\frac{X^5-1}{X-1}$$ and now i have to prove that is irreducible. 
The lineal change of variables are ok*(I don't know why) so I substitute $X$ by $X+1$ then I have:
$$\frac{(X+1)^5-1}{X}=\frac{X^5+5X^4+10X^3+10X^2+5X}{X}=X^4+5X^3+10X^2+10X+5$$
And now we can apply the Eisenstein criterion with p=5. So my polynomial is irreducible in $\mathbb{Q}$
Now let's prove that it has two different irreducible factors in $\mathbb{R}$
I've tryed this way: $X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D)$
and solve the system. But solve the system is quite difficult. Is there another way?

Every polynomial splits completely over the complexes. There are only three possibitilies:


*

*There are four real roots

*There are two real roots and one pair of complex conjugate roots

*There are two pairs of complex conjugate roots


The roots of the polynomial are fifth roots of unity other than $1$, so there are no real roots. Thus we are in the third case, and the factorization consists of two irreducible real quadratics.
