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$x=\sqrt 3 + i \sqrt 7$

I have come to the following polynomial: $x^4 + 8x^2+100=0$. By Eisenstein criterion, this pol is not irreducible in $\Bbb Q[x]$, but I don't know how to factor it further? Will be great if I can get some hints.

MathGeek
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GalaxyY
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    Why would that polynomial be not irreducible? The Eisenstein criterion only gives you a sufficient condition for irreducibility, not that the polynomial is reducible if said condition isn’t met. – Aphelli Nov 06 '21 at 13:31
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    I believe the minimal polynomial should be a 4th degree, why are you trying to factor it? – Alan Abraham Nov 06 '21 at 13:34
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    That polynomial is irreducible. – José Carlos Santos Nov 06 '21 at 13:34
  • You can easily finds the root of this polynomial and prove that indeed it's irreducible over $\mathbb Q$. – Surb Nov 06 '21 at 13:34
  • If you're hell-bent on trying to factor it, the rational root theorem tells you it has no rational roots, so if it factors, it must factor into two quadratics. Because it's monic you know the two quadratics would have to be of the form $(x^2+ax+b)(x^2+cx+d)$. If you try to solve this system of equations, you won't be able to with rational numbers, thus proving it's irreducible. – Merosity Nov 06 '21 at 14:33
  • A roadmap. 1) Show that $i\sqrt{21}\in\Bbb{Q}(x)$. 2) Therefore $K=\Bbb{Q}(i\sqrt{21})\subseteq L=\Bbb{Q}(x)$. 3) Conclude $[K:\Bbb{Q}]=2$ and that the real part of any element of $K$ is rational. 4) Conclude that $x\notin K$, and therefore $[\Bbb{Q}(x):\Bbb{Q}]\ge4$. 5) Conclude that the minimal polynomial has degree four at least. – Jyrki Lahtonen Nov 06 '21 at 15:23
  • Thanks everyone. Think I got enough tips to move forward. Appreciate all your input. – GalaxyY Nov 06 '21 at 15:34

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