Determine whether $f= 3x^3-3x+9$ is irreducible in $\mathbb Z[x]$ and if not, give a factorisation of f into a product of irreducibles.
What would be the best method to approach this? I understand Eisetein's is not applicable.
My idea so far:
No real valued roots from taking factors of 9.
So $3x^3-3x+9 = 3(x^3-x+3) = 3(x^2+ax+b)(x+c)=x^3+(c+a)x^2+(ac+b)x+bc$
$c+a=0$, $ac+b=-1$, $bc=3$
$a=-c \Rightarrow -c^2+b=1$ (by substitution)
Values of $b$ and $c$ must be $\pm 3, \pm1$ as $bc=3$, neither choice satisfies $-c^2+b=1$, so have no solution in $\mathbb Z$.
Then I'm not really sure where to go from here, or if I'm on the right lines in the first place!
Any help?
It is not irreducible since it is $3\cdot(x^3-x+3)$.
The factor $^3-x+3$ is irreducible, since it is irreducible modulo $2$, as it has no root i $\mathbf F_2$ (we might as well check it has no integer root, but it would be longer).
Of course the given polynomial is irreducible in $\mathbf Q[x]$, since $3$ is a unit in \mathbf Q$.