# Show that as a polynomial in $\mathbb{Q}[x]$, $P(x)$ is irreducible.

Let $$P(x)$$ be a polynomial in $$\mathbb{Z}[x]$$ of degree 5 such that $$P(1)=3$$ and $$P=(x-1)^5\bmod 3$$. Show that as a polynomial in $$\mathbb{Q}[x]$$, $$P(x)$$ is irreducible.

So here I find that $$P(x)=(x-1)^5+3=x^5-5x^4+10x^3-10x^2+5x+2.$$ How can I show this irreducible? Can I use Eisenstein's Criterion? How would I do that? For instance, if my prime is 5 then it divides 4 of the 6 terms but what about those other 2?

• Nope... not fixed... my question remains..... PLEASE take the time to post well-formed questions. Apr 23, 2020 at 0:10
• Sorry, I meant P(X). Apr 23, 2020 at 0:16
• Ummm.... this is getting tedious: $P(X)$ (in your comment) or $P(x)$??? PLEASE take the time to write a careful question. Apr 23, 2020 at 0:18
• It’s not $P(x)=(x-1)^5+3$; the fact that it is congruent to $(x-1)^5$ modulo $3$ tells you there is a polynomial $q(x)$ such that $P(x) = (x-1)^5 + 3q(x)$; you have no warrant to assert that $q(x)=1$. Apr 23, 2020 at 0:19

$$\!\bmod 3\!:\ p \equiv (x\!-\!1)^5\$$ so Eisenstein applies to $$f(x) := p(x\!+\!1) \equiv x^5\,$$ by $$\,3^2\nmid f(0)\! =\! p(1)\!=\! 3$$
Remark  This is the standard shifted Eisenstein criterion, e.g. see here for motivation.