P(x) = x^6-7x^3-8. Find all zeros of P, real and complex. Factor P completely.?

Answer 1

P(x) = x^6 - 7x³ - 8 = 0
(x³ + 1)(x³ - 8) = 0
(x + 1)(x² - x + 1)(x - 2)(x² + 2x + 4) = 0
(x + 1)(x - 2)(x² - x + 1)(x² + 2x + 4) = 0

x² - x + 1 = 0
x = [1 ± √(1 - 4)]/2 = [1 ± √(-3)]/2 = (1 ± i√3)/2

x² + 2x + 4 = 0
x = [-2 ± √(4 - 16)]/2 = [-2 ± √(-12)]/2 = (-2 ± i√12)/2
x = (-2 ± 2i√3)/2 = -1 ± i√3
____________

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

So the six roots of P(x), real and imaginary, are:

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

Answer 2

Factoring:
(x^3 - 8) (x^3 + 1) =
(x - 2)(x^2 + 2x + 4)(x^3 + 1) = 0
x = 2
use quad formula: -2 +/- sq rt of (4 - 4(1)(4) div 2(1)
-2 +/- 2i sqrt of 3 all div by2
and
x^3 + 1 = (x - 1)(x^2 + x + 1)
x = 1
use quad form to solve for x in x^2 + x + 1
-1 +/- sqrt of 1 - 4 div 2 = -1 +/- i sqrt 3 all div by 2

Answer 3

P(x) = (x^3-8)(x^3+1)

so we have P=0 when x^3 = 8 or x^3 = -1
x = 2 or x = -1

Answer 4

P(x)=x^6-7x^3-8
P(x)=(x-2)(x+1)(x^2 - x +1)(x^2 + 2x + 4)
P(x)=(x-2)(x+1)(x - ((1+sqrt(3)i)/2 ))(x - ((1-sqrt(3)i)/2 ))(x - (-1-sqrt(3)i))(x - (-1 + sqrt(3)i))