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Let $f(x)=x^{4}-10x^{2}-25\in \mathbb Q[x]$. I want to prove that $f(x)$ is a separable polynomial over $\mathbb Q$. I know that from the definition $f(x)$ is separable if none of the irreducible factors of $f(x)$ in $F[x]$ has a repeated root in a splitting field for $f(x)$ over $F$.

So for my example here $f(x)=x^{4}-10x^{2}-25$ $F=\mathbb Q$ and $F[x]=\mathbb Q[x]$

so do i just need to factorise $f(x)$ and show that the factors of $f(x)$ has no repeated roots in $\mathbb Q$?

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  • $\begingroup$ I think you can actually explicitly solve for the roots, and show that there are no repeated roots. Note that this is a quadratic equation in $x^2$. $\endgroup$ – fierydemon May 17 '18 at 20:15
  • $\begingroup$ $f$ is separable over $\Bbb{Q}$ iff it is relatively prime with its formal derivative $Df$ $\endgroup$ – Bumblebee May 17 '18 at 20:18
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No. You only have to prove $f(x)$ and its derivative have no common root in $\mathbf C$. So you just have to compute $\;\gcd(f, f')$ by Euclid's algorithm and check it is $1$.

Here, taking into account the particular value of $f(x)$, and $f'(x)=4x^3-20x=4x(x^2-5$, you only have to show that none of the irreducible factors of $f'(x)$ ($x$ and $x^2-5$ – divides $f(x)$). It amounts to checking neither $0$ nor $\sqrt 5$ is a root of $f$.

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  • $\begingroup$ Is there any particular method to do this, I’m not the greatest with long division unfortunately $\endgroup$ – Gibberish May 19 '18 at 3:50
  • $\begingroup$ The resultant of $f$ and $f'$ – but it's not simpler to compute (if $f$ has degree $d$, $\operatorname{Res}f,f'$ is a determinant of size $2d-1$. Why not a CAS? $\endgroup$ – Bernard May 19 '18 at 9:03
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Note that $f(x)=x^4-10x^2-25=(x^2-5)^2-50$ has four distinct zeros given by $$\pm\sqrt{5\pm\sqrt{50}}.$$

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