#1**+1 **

I think that is going to be pretty ugly....Are you sure it isn't 16x^4** - **1 ?

ElectricPavlov Nov 4, 2018

#2**+1 **

ok so basically the zeros of this are

\(x =\dfrac 1 2 e^{\dfrac{i(\pi + 2\pi k)}{4}},~k=0,1,2,3\)

and then what we do is pair up the conjugates and form

\(\left(x - \dfrac 1 2 e^{i\pi/4}\right)\left(x - \dfrac 1 2 e^{-i\pi/4}\right) \left(x - \dfrac 1 2 e^{3\pi/4}\right)\left(x - \dfrac 1 2 e^{-3i\pi/4}\right)\)

which becomes

\(\left(x^2 - \dfrac{x}{\sqrt{2}}+\dfrac 1 4\right)\left(x^2+\dfrac{x}{\sqrt{2}}+\dfrac{1}{4}\right)\)

.Rom Nov 4, 2018