How do I show $ x^3-tx^2+(t-3)x+1 $ is irreducible? (Shanks' simplest cubic) Let $K = {\mathbb Q}[t]$. Show $x^3 - tx^2 + (t-3)x + 1$ is irreducible in $K[x]$.
I tried substitution  with $x-t$ and other things, hoping to use Eisenstein's criterion to finish the job. But I have not made much progress. Can I get a hint?
Thank you.
 A: I’ll use freely the fact that $\mathbb Q[t]$ and $\mathbb Q[x]$ are unique factorization domains. Rewrite your polynomial as an element of $\mathbb Q[x][t]$, giving the linear (in $t$) polynomial $(x-x^2)t + (x^3-3x+1)$. Any factorization would have to be linear (in $t$) times a constant, where this last word means in $\mathbb Q[x]$. But that would say that $x-x^2$ and $x^3-3x+1$ had a common factor, which they don’t (except for genuine constants in $\mathbb Q$, of course).
A: Suppose there exists $\,p(t)\in K\,\,\,s.t.\,\,\,p(t)^3-tp(t)^2+(t-3)p(t)+1=0\,$ .
Hint: since $\,p(t)\in\Bbb Q[t],\,$ , the above implies $\,t\,$ is algebraic over the rationals...
A: Did I hear Eisenstein's criterion?
Suppose for a contradiction that $f_t(x)=x^3-tx^2+(t-3)x+1$ is reducible in $(\mathbb{Q}[t])[x]$, i.e., $f_t(x)$ factors as $f_t(x)=g_t(x)h_t(x)$ where $g_t,h_t\in (\mathbb{Q}[t])[x]$ are non-constant as a function of $x$. Then $f_{t_0}(x)$ must be reducible for all values of $t_0\in\mathbb{Q}$ (because $f_{t_0}(x)=g_{t_0}(x)h_{t_0}(x)$) as a polynomial in $\mathbb{Q}[x]$. In particular, when $t_0=0$, the polynomial 
$$f_0(x)=x^3-3x+1$$
is reducible. But Eisenstein's criterion (for $p=3$) says that $f_0(x+2)=\tilde{f}_0(x)= x^3 + 6x^2 + 9x + 3$ is irreducible, and we have reached a contradiction.
