As stated, I am trying to prove that $10X^6-15X^2+7$ is irreducible in $\mathbb{Q}[X].$

I have been given the hint to compare the above polynomial with $7X^6 - 15X^4+10.$

I know that $7X^6 - 15X^4+10$ is irreducible in $\mathbb{Q}[X]$ because of eisenstein's criterion for $p=5$, but I have no idea how one could prove that $10X^6-15X^2+7$ is irreducible because $7X^6 - 15X^4+10$ is irreducible. Any help would be appreciated

  • 2
    $\begingroup$ WA tells us that $10X^6-15X^2+7$ is irreducible mod 11. $\endgroup$
    – lhf
    Jul 2, 2018 at 21:21

2 Answers 2


HINT: If $f(x)$ is irreducible, so is $x^nf(1/x)$, where $n = \deg f$


Suppose that $10X^6-15X^2+7=P(X)Q(X)$, where $P(X)$ and $Q(X)$ are non-constant polynmials, whose degrees are $m$ and $n$ respectively. Then $m+n=6$ and$$X^6P\left(\frac1X\right)Q\left(\frac1X\right)=7X^6-15X^4+10.\tag1$$But $X^6P\left(\frac1X\right)Q\left(\frac1X\right)=X^mP\left(\frac1X\right)X^nQ\left(\frac1X\right)$ and therefore $(1)$ expresses $7X^6-15X^4+10$ as the product of two non-constant polynomials.


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