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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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    Can this observation help? Your polynomial can be rewritten as $(1+x)^5 -2x -4$.2011-11-20
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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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    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!2011-11-20
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    He probably recognized that $p(x)$ is very close to $(x+1)^5$...2011-11-20
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    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!2011-11-20
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    Ah, that's fine too. I might try to read the answers to [this question](http://math.stackexchange.com/questions/23874/motivation-for-eisenstein-criterion) later on. Thanks to N. S. as well.2011-11-20
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    @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.2011-11-20
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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