# Why is $x^5+5x^4+10x^3+10x^2+7x+5$ irreducible over $\mathbb{Q}[x]$?

I am given the polynomial

$x^5+5x^4+10x^3+10x^2+7x+5$,

and shall show that it is irreducible over $\mathbb{Q}[x]$. The only thing that we have been introduced until now is Eisenstein's criterion, and it would almost work here. So is there any trick that can be done on the coefficient $7x$ to apply Eisenstein's criterion, or do we need something else here?

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Can this observation help? Your polynomial can be rewritten as $(1+x)^5 -2x -4$. – user38268 Nov 20 '11 at 19:44
@Benjamin: You mean $(1+x)^5 +2x +4$, I think. – TonyK Jan 24 '12 at 16:25

Letting your polynomial be $$p(x)=x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 7 x + 5,$$ take a look at $p(x-1)$ and apply Eisenstein there. Then show that $p(x)$ is irreducible if and only if $p(x-1)$ is.

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Could you say a few words about how you came up with this subsitution? I tried a few shifts, but I'm wondering whether there's some trick for recognizing the right one. Nice answer! – Dylan Moreland Nov 20 '11 at 19:17
He probably recognized that $p(x)$ is very close to $(x+1)^5$... – N. S. Nov 20 '11 at 19:18
I'm afraid I don't have a trick; I used Mathematica and by luck this was the first shift I tried. Whether there is such a trick would make for an interesting question in its own right though! – Zev Chonoles Nov 20 '11 at 19:20
Ah, that's fine too. I might try to read the answers to this question later on. Thanks to N. S. as well. – Dylan Moreland Nov 20 '11 at 19:22
@N.S. I did recognize that, but I don't think I consciously realized that it would make $x-1$ a good shift to try; now I see the connection. – Zev Chonoles Nov 20 '11 at 19:23

Observe that $$p(x)=(x+1)^5+2(x+1)+2.$$ It suggests you should change variable to $$y=x+1,$$ so that $$\phi(y)=p(y-1)=y^5+2y+2.$$

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Which is (strictly) contained in Zev's answer. – Did Jan 24 '12 at 17:22