Show that the polynomial $x^4 - 4x^3 + 6$ is irreducible in $\mathbb{Z}[x]$

Question

Show that the polynomial $x^4 - 4x^3 + 6$ is irreducible in $\mathbb{Z}[x]$.

Theorem : Let $A$ an integral domain and $I$ a proper ideal of $A$. If $f(x) \not \equiv a(x)b(x) \pmod I$ for any polynomials $a(x)$, $b(x)$ $\in A[x]$ of degree $\in [1, \deg(f))$, then $f(x)$ is irreducible in $A[x]$

I think I have to use the previous theorem, but I don't know how to use it. Is anyone could help me at this point?

Answer 1

Reduce the polynomial modulo $4$ to get the polynomial $x^4+2\in\mathbb{Z}/4\mathbb{Z}$. Observe that this polynomial has no roots in $\mathbb{Z}/4\mathbb{Z}$. Argue that the polynomial cannot be factored as the product of two irreducible quadratic polynomials in $\mathbb{Z}/4\mathbb{Z}$ and hence is irreducible there. Conclude using the theorem.

Answer 2

May be you should solve for $a,b$ in

$x^4-4x^3+6=(x^2+ax+1)(x^2+bx+6)\Rightarrow a+b=-4; 7+ab=0; 6a+b=0$

$x^4-4x^3+6=(x^2+ax-1)(x^2+bx-6)\Rightarrow ??$

$x^4-4x^3+6=(x^2+ax+2)(x^2+bx+3)\Rightarrow ??$

$x^4-4x^3+6=(x^2+ax-2)(x^2+bx-3)\Rightarrow ??$