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The problem suggests using a suitable change of variable of the form $y=ax+b$ and using Eisenstein's criterion to show that it's reducible in $\mathbb{Q}[x]$. So, I have several doubts, first of all, isn't there a problem when doing a change of variable of the form $y=ax+b$ when $a\not=1$ (Since $x=\frac{y-b}{a}$ and this may not be in $\mathbb{Z}$, which I need in order to use the irreductibility criterion). On the other hand, in case there's no problem with the change of variable, how can I choose $a, b$ in an intelligent way?. I tried thinking what $2(ax+b)^4+15(ax+b)^2+10$ would be in it's expanded form in order to know the expression for each coefficient in terms of $a$ and $b$ but I have not achieved anything important. How can I choose $a,b$ wisely for this particular case? Is there a way to do so for a most general case (any irreductible polynomial)? Thanks in advance

2 Answers2

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You don't actually need a change of variables. Eisenstein's criterion applies directly with $p=5$.

sbares
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  • This would have been my answer, except that as usual, I would have stretched it out into a paragraph, saying something about everything being all right if the leading coeff is indivisible by $p$. Anyway, $+1$. – Lubin Sep 12 '18 at 00:59
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Eisenstein's criterion can be applied to a polynomial $f(x)$, with replacing $x$ by $ax+b$. There is no problem with integer coefficients, because we are asking irreducibility for $\Bbb{Q}[x]$ with rational coefficients and have Gauss Lemma for integral coefficients. So try replacing $x$ by $x$ itself first, and then some other examples helping you to find $a$ and $b$. What about $x+5$ ?

If you want to know more methods in general, just search on this site. For example, the following link has several answers which you can also use for your polynomial:

Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$

Another solution: It is easy to see that $2x^4+15x^2+10$ is irreducible over $\Bbb{F}_7$. Since the leading coefficient is not divisible by $7$, it is also irreducible over $\Bbb{Z}$.

Dietrich Burde
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