factorising polynomials I got stuck on my algebra homework. 
First of all, can you check if the following statements are correct?

The polynomial $f=3X^8+6X^4+2$ is irreducible in $\mathbb{Z}[X]$ because $f \in \mathbb{F}_3$ equals $2$, which is irreducile. The polynomial is primitive, Gauss's lemma tells us that $f$ is irreducible in $\mathbb{Q}[X]$ as well.
The irreducible factorisation of $4X^6+8X^3+4$ in $\mathbb{Z}[Xb]$ is the same as in $\mathbb{Q}[X]$ and equals$4(X+1)^2(X^2-X+1)^2$

Now I'll show you some polynomials I got stuck on, With my research effort. Could you bring me a hint (just a hint) to factorise each of these polynomials in $\mathbb{Z}[X]$ and in $\mathbb{Q}[X]$?



*

*$X^5-2X^4+X^3-2X^2-2$. I didn't see if this was reducible or not. It has a root, but I don't know if it's rational. It's not an Eisenstein polynomial.

*$X^6+X^3+1$. This polynomial has no zeros, so it's irreducible in $\mathbb{Z}[X^3]$. That's all I knew.

*$X^3+3X^2+4X+5$. Again, this polynomial is primitive. What's next?



As you see, I've just very little experience with this. I hope you can help me.
 A: The first claim is not fully satisfactory. For example $9X^8-1$ also equals $2$ when reducted modulo $3$, but it factors as $(3X^4-1)(3X^4+1)$ in $\mathbb Z[X]$!
The second factorization is essentially correct, but strictly speaking $4$ is a unit in $\mathbb Q[X]$, but factors into irrducibles as $4=2\cdot 2$ in \mathbb Z[X]$.
You argue that $X^5-2X^4+X^3-2X^2-2$ has a root - in $\mathbb R$, I presume, because of the odd degree. But that does not help us at all with factorization in $\mathbb Q[X]$. As a matter of fact, the polynomial is irreducible, but why? If all else fails, check if a linear factor is possible and check if a quadratic factor is possible. A quadratic factor would have to look like $X^2+aX+b$ with $a\in\mathbb Z$ and $b\in\{-2,-1,1,2\}$
$X^6+X^3+1$ is irreducible because its six complex roots are also roots of $X^9-1=(X^3-1)(X^6+X^3+1)$, i.e. nineth roots of unity and we know that there are six primitive such roots (the non-primitive ones being the roots of $X^3-1$).
If $X^3+3X^2+4X+5$ could be factored, at least one factor would be linear, so one of $X\pm1$, $X\pm 5$ (why?).
