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?

let $$P(x)=x^4+x^3+x^2+x^1+1$$ 
We know if $x=\frac{a}{b}$ is root of $P(x)$ then $b|1\,$ , $\,a|1$. In the other words $a=\pm 1 $ and $b=\pm 1 $ but $P(1)=5$ and $P(-1)=1$, thus we let
$$P(x)=(x^2+ax+b)(x^2+cx+d)$$
as a result
 \begin{align}
  & bd=1 \\ 
 & ad+bc=1 \\ 
 & b+d+ac=1 \\ 
 & a+c=1 \\ 
\end{align}
This system has not solution in $Q$ because
$$d(ad+bc)=d\times\,1\to\,ad^2+c=d$$
 On the other hand $\,c=1-a$ thus 
$$ad^2+1-a=d\to\,a(d^2-1)=d-1$$ 
This implies $d=1$ or $ad+a=1$. If $d=1$ then $\left\{\begin{matrix}
   a+c=1  \\
   ac=-1  \\
\end{matrix}\right.$ that this system has not rational roots . If $\,ad+a=1\,$ then $a=\frac{1}{d+1}=\frac{b}{b+1}$ as a result
$$b+d+ac=1\to b+\frac{1}{b}+\frac{b}{b+1}\left(1-\frac{b}{b+1}\right)=1$$
we have
$$\frac{(b+1)^2}{b}+\frac{b}{(b+1)^2}=-1$$ This equation has not solution in $\mathbb{R}$
