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# Factorization w/Real Coefficients

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Factor 16x4+1

Nov 3, 2018

#1
+18360
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I think that is going to be pretty ugly....Are you sure it isn't  16x^4 - 1   ?

Nov 4, 2018
#2
+5088
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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