State how many imaginary and real zeros the function has. f(x) = x4 - 8x3 + 17x2 - 8x +16

  x^4 - 8x^3 + 17x^2 - 8x + 16 = 0
x^4 - 8x^3 + 16x^2 + x^2 - 8x + 16 = 0
x^2(x^2 - 8x + 16) + (x^2 - 8x + 16) = 0
(x^2 + 1)(x^2 - 8x + 16) = 0
(x^2 + 1)(x - 4)^2 = 0
x^2 + 1 = 0 or x - 4 = 0
x^2 = -1 or x = 4
x = +/- i or x = 4
Technically there are 2 imaginary solutions and 1 real solution. However, since the real solution occurs twice (multiplicity 2), you could say there are 2 real solutions (although they're the same). So the second answer would apply.