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Factor 16x4+1 

 Nov 3, 2018
 #1
avatar+18360 
+1

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

 Nov 4, 2018
 #2
avatar+5088 
+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

 

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

.
 Nov 4, 2018
edited by Rom  Nov 4, 2018

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