Find all real and complex solutions of the polynomial equation. x^3-2x^2+4x-8=0 a. -2, 2i, -2i b. 2, 2i, -2i c. 2, -2, -2i d. 2, -2, 2i, -2i

First we want to solve by factoring. So take x^3 - 2x^2 + 4x - 8 = 0, add parenthesis > (x^3 - 2x^2) + (4x - 8) 
Now factor out x^2 from x^3 - 2x^2 which is x^2(x - 2), factor out 4 from 4x - 8 which is 4(x - 2) > Now we have > x^2(x - 2) + 4(x - 2) [Factor out the common term of (x - 2)] > (x - 2)(x^2 + 4) = 0
Okay now we want to apply the zero factor principle.
x - 2 = 0 > Add 2 to both sides > x - 2 + 2 = 0 + 2 > x = 2.

Now we have to plug into the quadratic equation where a = 1, b = 0, c = 4
We get -0 + sqrt0 - 1 * 4 * 4/1 * 2 after applying the rule of 0^a = 0 to -0 + sqrt0^2  - 4 * 1 * 4/2 * 1.
Multiply the numbers now...
sqrt0 - 4 * 1 * 4 > sqrt0 - 16 > subtract > sqrt-16. Apply the rad rule of..

sqrt-1, sqrt-16, Apply the imaginary rule sqrt-1 = i > sqrt16i
Now we have sqrt16i/2, so we want to factor 16 which is 2^4 because 2 * 2 * 2 * 2 = 16. Assuming a > 0 we want to apply 
2^2i/2 > cancel the common factor of 2 > 2i
For the next part, do the exact same steps except when you get to -sqrt16i/2, apply the fraction rule > 
To get -2i.
Therefore our final solutions are, (2, 2i, -2i)