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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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    @Benjamin: You mean $(1+x)^5 +2x +4$, I think.2012-01-24

2 Answers 2

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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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    One way to see the trick would be to notice that $x^5 + 5x^4 + 10x^3$ are the three first terms of the binomial expansion of $(1+x)^5$, thus one can expect to write $p$ as $(1+x)^5 + $ terms of low degree. Magically looking at $p(x-1)$ works, but that's it. If it didn't, we would probably still be screwed.2011-11-20
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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.2012-01-24