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.