State whether $x^5-5x^4+10x^3-7x^2+8x-4$ is irreducible or not I have tried everything in my knowledge and no, I cannot state it.
I have tried a factorizor online which tells me that it is not factorizable hence irreducible. But I cannot reason why.
I looked at Eisenstein's criteria but obviously, there is no prime $q$ that fits the criteria so this is useless.
I then tried reducibility via modulo reduction, and this should give me the options to test irreducibility up to mod $8$ since that is the largest coefficient in the polynomial...yes? But every mod arrives at the polynomial being reducible...so it basically fails to tell me that it is irreducible.
For mod 2, I get 0 as a solution to the reduced polynomial so that means I can factor it out with $x$.
Similarly, mod 3 says 1 is a solution so $(x-1)$ should be a solution.
In a similar fashion, I get mod 4,5,6,7 to have solutions 0,3,4,6.
Am I missing anything? This is the best I can do, nothing more assures me irreducibility at all. Ideas please...?
 A: Let $\phi(x)$ denote your polynomial.  Then we note that $$\phi(x+1)=x^5+3x^2+9x+3$$  and we can invoke Eisenstein's criterion.
A: Let $x = y+1$.  Then this equation becomes:
$y^5+3y^2+9y+3$.  Use Eisenstein to show this is irreducible.  Therefore, the original is also irreducible.
A: One option is to reduce the given polynomial modulo $11$, in which case it factors (over $\Bbb F_{11}$) as
$$(x - 5)(x^4 - x^2 - x - 3).$$
So, if the polynomial is reducible over $\Bbb Q$, it has one linear factor and one irreducible quartic factor there.
On the other hand, checking the short list, $\pm 1, \pm 2, \pm 4$, of candidates given by the Rational Root Test shows that the polynomial has no rational roots and hence no linear factors. (In fact, since the signs of the polynomial are alternating, all of its real roots are positive, so we need only check $+1, +2, +4$.) Thus, the polynomial is irreducible.
(Alternatively, the given polynomial is irreducible altogether modulo $17$, but this is surely even less pleasant to check by hand than the irreducibility modulo $11$ of the above quartic.)
