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?

A different route to the factorization over the reals (obviously the end result is same as in Egreg's post, but I give the factors explicitly).
Let $p(x)$ be your polynomial. By a direct calculation we see that
$$
(x^2+\frac x2+1)^2=x^4+x^3+\frac94x^2+x+1=p(x)+\frac54 x^2.
$$
This calculation is aided by palindromic symmetry of both $p(x)$ and this quadratic. Anyway, this gives us the factorization
$$
\begin{aligned}
p(x)&=(x^2+\frac x2+1)^2-(\frac{\sqrt5}2\,x)^2\\
&=(x^2+\frac{1-\sqrt5}2\, x+1)(x^2+\frac{1+\sqrt5}2\,x+1)
\end{aligned}
$$
by the usual
$$
a^2-b^2=(a-b)(a+b)
$$
formula.
So the Golden ratio (not surprisingly given that the zeros are vertices of a regular pentagon) makes an appearance.
