polynomials/complex no's q (1 Viewer)

[bally24]

from cambridge 4.4: show that the zeroes of P(x)=x⁴-x²+1 are included in the zeroes of x⁶+1. Hence factorise P(x) over R.

first part is fine, i factorised it as a cubic. but how do you find roots, not including x=1, x=-1, of x^6+1? since it is an even power and so 1 is not a solution. i dunno im just really confused about it. thank for your help.

edit: i know it used complex roots of unity but i dont understand how to do that.

 

[pLuvia]

x⁶+1=0
x⁶=-1
x⁶=cis(pi+2kpi)
...

Can you work it out from there?

 

[bally24]

k is 0, 1, 2, 3.... right? then won't you get the solutions just as cis(pi), cis(3pi), cis(5pi) etc? is that right? lol im probably been really dumb here...

 

[pLuvia]

k is 0, 1, 2, 3.... right? then won't you get the solutions just as cis(pi), cis(3pi), cis(5pi) etc? is that right? lol im probably been really dumb here...

k=0,+1,+2,3 easier this way but your method is fine.
Yep that looks correct. Then you can easily turn the solutions back into cartesian form

 

[bally24]

thats great...thanks!

 

[pLuvia]

thats great...thanks!

Actually sorry, misread your answers they should be cis(pi/6), cis(pi/2), cis(5pi/6) ... Since x=cis[(pi+2kpi)/6]

 

[Riviet]

The reason why taking k=0,+1,+2, etc. is easier is because you won't get any improper fractions which might otherwise need converting, ie subtracting/adding 2pi to get it into the range from -pi to pi.