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: Finding the complex roots of the polynomial is easy: if $\varphi=2\pi/5$, the roots are
$$
r_1=e^{i\varphi},\quad
r_2=e^{2i\varphi},\quad
r_3=e^{3i\varphi}=\bar{r}_2\quad
r_4=e^{4i\varphi}=\bar{r}_1
$$
and so the factorization over the reals is
$$
(X^2-(r_1+\bar{r}_1)X+1)((X^2-(r_2+\bar{r}_2)X+1).
$$
What you want to prove is that this factorization is not over $\mathbb{Q}$, thereby deducing that the polynomial is irreducible over $\mathbb{Q}$.
The procedure is standard: let $r$ be any root of the polynomial; then
$$
r^2+r+1+\frac{1}{r}+\frac{1}{r^2}=0
$$
and so
$$
\left(r+\frac{1}{r}\right)^2+\left(r+\frac{1}{r}\right)-1=0
$$
Since the polynomial $X^2+X-1$ has no rational root, you have proved that
$$
r_1+\bar{r}_1=r_1+\frac{1}{r_1}
$$
is not rational.

If $p$ is prime, then $X^{p-1}+X^{p-2}+\dots+X+1$ is irreducible over $\mathbb{Q}$. Write it as
$$
\frac{X^p-1}{X-1}
$$
and substitute $X=Y+1$. You'll see that Eisenstein applies.
A: 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 $\mathbb{R}$
A: 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.
A: 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.
