Um die Brüche wegzukriegen, multiplizieren wir den Term mit ${\displaystyle {}256}$ und erhalten

${\displaystyle {\left(4x-{\left({\sqrt {3}}+1\right)}\right)}^{2}{\left(4x-{\left(-{\sqrt {3}}+1\right)}\right)}^{2}-32x-4.}$

Es ist

${\displaystyle {}{\left(4x-{\left({\sqrt {3}}+1\right)}\right)}^{2}=16x^{2}-8{\left({\sqrt {3}}+1\right)}x+{\left({\sqrt {3}}+1\right)}^{2}\,}$

und

${\displaystyle {}{\left(4x-{\left(-{\sqrt {3}}+1\right)}\right)}^{2}=16x^{2}-8{\left(-{\sqrt {3}}+1\right)}x+{\left(-{\sqrt {3}}+1\right)}^{2}\,.}$

Deren Produkt ist

${\displaystyle {}{\begin{aligned}{\left(16x^{2}-8{\left({\sqrt {3}}+1\right)}x+{\left({\sqrt {3}}+1\right)}^{2}\right)}{\left(16x^{2}-8{\left(-{\sqrt {3}}+1\right)}x+{\left(-{\sqrt {3}}+1\right)}^{2}\right)}&=px^{4}+qx^{3}+rx^{2}+sx+t\\\end{aligned}}}$

und die Koeffizienten sind

${\displaystyle {}p=256\,.}$

${\displaystyle {}q=-128{\left({\sqrt {3}}+1-{\sqrt {3}}+1\right)}=-256\,,}$

${\displaystyle {}{\begin{aligned}r&=16{\left({\left({\sqrt {3}}+1\right)}^{2}+{\left(-{\sqrt {3}}+1\right)}^{2}\right)}+64\cdot ({\sqrt {3}}+1)(-{\sqrt {3}}+1)\\&=16\cdot 8+64\cdot (-2)\\&=0,\end{aligned}}}$

${\displaystyle {}s=-8({\sqrt {3}}+1)(-{\sqrt {3}}+1){\left({\sqrt {3}}+1-{\sqrt {3}}+1\right)}=-8(-2)\cdot 2=32\,}$

und

${\displaystyle {}{\begin{aligned}t&={\left({\sqrt {3}}+1\right)}^{2}{\left(-{\sqrt {3}}+1\right)}^{2}\\&=4.\end{aligned}}}$

${\displaystyle {}x^{4}-x^{3}}$