What is "prime factorisation" of polynomials? I have the following question:

Find the prime factorization in $\mathbb{Z}[x]$ of $x^3 - 1, x^4 - 1, x^6 - 1$ and $x^{12} - 1$. You will need to check the irreducibility in $\mathbb{Z}[x]$, of three quadratic polynomials and of one quartic. In the case of the quadratic, you will need to check that it has no integer zeros and does not factorize as a product of two quadratics with integer coefficients.

For starters, what does prime factorization mean? If we look at $x^3 - 1$ as an example, I can see that this can't be factorized any further, so it is irreducible in $\mathbb{Z}[x]$. Does that mean there is no prime factors for $x^3 - 1$?
If we then look at $x^4 -1 = (x^2 - 1)(x^2 + 1)$. Here, we see that it is irreducible and so would it simply be that the prime factorization of $(x^4 -1 ) = (x^2 + 1)(x - 1)(x + 1)$?
 A: For quadratics and cubics, irreducibility is easy to test, for if there are no rational roots, the polynomial must be irreducible. And the polynomials $x^2+x+1$  and $x^2-x+1$ don't even have real roots. In more complicated cases, the Rational Roots Theorem could be helpful.
Let's deal with $x^{12}-1$. This immediately factors as $(x^6-1)(x^6+1)$. We leave further decomposition of $x^6-1$ to you. For $x^6+1$, note that it is equal to $(x^2+1)(x^4-x^2+1)$. 
We would like to show that $x^4-x^2+1$ is irreducible. It certainly cannot be written as a linear polynomial with coefficients in $\mathbb{Z}$ times a cubic, since $x^4-x^2+1=0$ has no rational roots. Can we express $x^4-x^2+1$ as a product of quadratics with integer coefficients? 
If we can, without loss of generality we would have that the decomposition has shape $(x^2+ax+b)(x^2+cx+d)$. Expand. Since $x^4-x^2+1$ has no $x^3$ term, we must have $a+c=0$, so $c=-a$.
Also, we must have $b=1$, $d=1$ and $b=-1$, $d=-1$. This gives the two possibilities (i) $(x^2+ax+1)(x^2-ax+1)$ and (ii) $(x^2+ax-1)(x^2-ax-1)$. 
We check there is no factorization of type (i). For the coefficient of $x^2$ in the product $(x^2+ax+1)(x^2-ax+1)$ is $2-a^2$. This cannot be $-1$ for integer $a$. 
Similarly, one can show (ii) can't work. 
Remark: We want to express each polynomial as a product of irreducibles. The factorization $x^4-1=(x-1)(x+1)(x^2+1)$ is such a factorization. It is not quite unique. For example we also have $x^4-1=(-x+1)(x+1)(-x^2-1)$, and two other obvious variants. But the decomposition is essentially unique, so giving one decomposition is enough. 
A: The prime factorization is the factorization into irreducible polynomials. For example, since $x^3-1$ can be factorized as $(x-1)(x^2+x+1)$ and both these are irreducible, the prime factorization of $x^3-1$ in $\mathbb Z[x]$ of $x^3-1$ is $(x-1)(x^2+x+1)$.
