Find all unique zeros of the function fix) 8x- 64x + 160x- 128-

Lets divide by 8. My god, those coefficients are large.

x^3 - 8x^2 + 20x - 16 = 0

Possible integral roots are: +1, -1, +2, -2, +4, -4, +8, -8

If x = 2, f(x) = 8 - 32 + 40 - 16 = 0

Hence, x-2 is a factor.

Use synthetic division to get: x^2 - 6x + 8 = (x-4)(x-2).

Therefore, there are two integral roots: +2 and +4.

Step-by-step explanation:

the correct answer is 1: 2

step-by-step explanation:

hope this your welcome : )