How to go about proving that $-1+(x-4)(x-3)(x-2)(x-1)$ is irreducible in $\mathbb{Q}$?

Question

How do you show that $-1+(x-4)(x-3)(x-2)(x-1)$ is irreducible in $\mathbb{Q}$?

I don't think you can use the eisenstein criterion here

Answer 1

Using the Gauss lemma as suggested, note that $1,2,3,4$ are clearly not roots of $p(x)$. Also, when $x\leq 0$ or $x\geq 5$ we have $p(x)\geq -1+24>0$. So there are no integer roots. So $p$ is irreducible over $\mathbb{Z}$ and hence $\mathbb{Q}$.

Answer 2

Actually, the obvious generalization is also true. Let $P$ and $Q$ be polynomial factors, so that the given expression equals $PQ$. Then $PQ=-1$ at each of the integer values, $1,2,3,4$ for this case. So $P,Q=\pm1$ at each of these integers, and whenever $P$ equals one, $Q$ equals minus one, and vice versa. But these are more than enough values to determine each polynomial, so $P=-Q$. But this is impossible because the coefficient of the highest order term in the product must be plus one.

Answer 3

Another argument, special to this polynomial.

Any number of arguments (including multiplying the thing out) show that the polynomial is $\equiv X^4-2\pmod5$. Thus in characteristic $5$, any root is a $16$-th root of unity. But the smallest field of characteristic $5$ whose cardinality is $\equiv1\pmod{16}$ is $\Bbb F_{625}$. Thus the polynomial is already irreducible over $\Bbb F_5$, thus over $\Bbb Z$, thus over $\Bbb Q$.