# Factorisation of a polynomial over $\mathbb Q(i,\sqrt 5)$

I'm trying to factorize a polynomial over $\mathbb Q(i,\sqrt 5)$:

$f(x)= x^4 -4x^2 - 5$.

I've factorised it thus:

in the form $(x^2 + ax + b)(x^2 + cx + d)$ where $b =d = \sqrt{-5}$ and $a=-c=\sqrt{2\sqrt{-5}+4}$ and I've then factorized each of these quadratics further to the forms $(x+a)(x+b)$ so that the overall factorization of the original polynomial is $(x-i)(x+i)(x-\sqrt 5)(x+\sqrt 5)$.

When I multiply the above out to get quadratics again to check my answers, the $x$ coefficient I get is $(i+\sqrt 5)$. Could someone please tell me why $(i+\sqrt 5)$ is equal to $\sqrt 2{\sqrt{-5} + 4}$? Or have I done this factorisation in an incorrect way, though multiplying out everything seems to work out fine.

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$x^4-4x^2-5=(x^2-2)^2-9=((x^2-2)+3)((x^2-2)-3)=(x^2+1)(x^2-5)=(x+i)(x-i)(x+\sqrt‌​{5})(x-\sqrt{5})$. –  Andrea Mori Apr 22 '12 at 9:59
Also, $(\sqrt{5}+i)^2=5+2i\sqrt{5}-1=4+2\sqrt{-5}$. –  Andrea Mori Apr 22 '12 at 10:03
Of course..! Thank you for the point out :) –  user29553 Apr 22 '12 at 10:08
Hint $\$ Factor $\rm\: X^2 - 4\:X - 5\:$ then put $\rm\:X = x^2$
Its more obvious for integers, e.g. with $\rm\:X = 10^n$
$\rm\qquad\begin{eqnarray} 5\!\!&\cdot&\!\! 11 &=& 55 \\ 95\!\!&\cdot&\!\! 101 &=& 9595 \\ 995\!\!&\cdot&\!\! 1001 &=& 995995 \\ 9995\!\!&\cdot&\!\! 10001 &=& 99959995 \\ & \cdots & & &\quad \cdots \\ (10^n\!-\!5)\!\!&\cdot&\!\! (10^n\!+\!1)\:\! &=&\:\! 10^{2n}\!-4\cdot10^n\! -\! 5 \end{eqnarray}$