# 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?

• 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.
• He probably recognized that $p(x)$ is very close to $(x+1)^5$... – N. S. Nov 20 '11 at 19:18
• @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 $$\phi(y)=p(y-1)=y^5+2y+2.$$