x5−8x4+28x3−56x2+64x−32=0

Answer 1

tried: ±1 , ±2 and found
x = 2 to be a triple root.

successive synthetic divisions shows

x^5−8x^4+28x^3−56x^2+64x−32=0
(notice how we show exponents)

x=2 to be triple root

so
x^5−8x^4+28x^3−56x^2+64x−32=0
(x-2)^3 (x^2 - 2x + 4) = 0

our remaining quadratic has no real roots
x = (2 ± √-12)/2
= 1 ± √-3
= 1 ± i√3

so x = 2 , 1 ± i√3

verification:
http://www.wolframalpha.com/input/?i=x%5E5−8x%5E4%2B28x%5E3−56x%5E2%2B64x−32%3D0