Find all zeros of the polynomial

Question

Find all zeros of the polynomial 

P(x)=x⁶-1

x1=, x2= with x1<x2

x3= ()+()i with both negative real and imaginary parts

x4=()+()i with negative real part and positive imaginary part

x5= ()+()i with positive real part and negative imaginary part, 

x6=()+()i with with both positive real and imaginary parts. 

Answer 1

Hi Dalia,

 

P(x)=x⁶-1=(x³)²-(1)²

               =(x³-1)(x³+1)

 

               =(x-1)(x²+x+1)(x+1)(x²-x+1)

 

x-1=0

x=1

x+1=0

x=-1

 

x²+x+1=0

 

x=-1±√1-4(1)(1)

         2(1)

 

  =(-1±√-3)/2

 

 =-1/2±i√3/2

 

x²-x+1=0

 

x=-(-1)±√(-1)²-4(1)(1)

            2(1)

 

 =(1±i√3)/2

 

=1/2±i√3/2

 

The zeros of the polynomial are 1, -1, -1/2-i√3/2, -1/2+i√3/2, 1/2-i√3/2, 1/2+i√3/2

Answer 2

P(x)=x^6-1 = (x^3+1)(x^3-1)

 

P(x) = (x+1)(x^2-x+1)(x-1)(x^2+x+1)

 

x^2±x+1=0

 

x^2+2(±1/2)x+(±1/2)^2-(±1/2)^2+1=0

 

(x±1/2)^2 - 1/4 + 4/4 = 0

 

(x±1/2)^2 = - 3/4

 

x±1/2 = ±i√(3)/2

 

x = ±1/2 ± i√(3)/2

 

The 6 zeros of P(x) are:

±1

+1/2 ± i√(3)/2

-1/2 ± i√(3)/2

 

check:

 

P(x)=
(x+1/2 + i√(3)/2)(x+1/2 - i√(3)/2)
(x-1/2 + i√(3)/2)(x-1/2 - i√(3)/2)
(x+1)(x-1) =

(x^2 + 1/2 x - x i√(3)/2
        + 1/2 x          + 1/4 - 1/2 i√(3)/2
                   + x i√(3)/2 + 1/2 i√(3)/2 + 3/4)
(x^2 - 1/2 x - x i√(3)/2
        - 1/2 x        + 1/4 + 1/2 i√(3)/2
                  + x i√(3)/2 - 1/2 i√(3)/2 + 3/4)
(x+1)(x-1)=

(x^2 + x + 1)
(x^2 - x + 1)
(x+1)(x-1)=

(x-1)(x^2 + x + 1)
(x+1)(x^2 - x + 1)
=

(x^3 - 1)(x^3 + 1) =

x^6 - 1 √