Which are the real zeros of this function? f(x) = x3 – 6x2 – 16x A. 0, 1, 8 B. 8, -1, 4 C. -4, 0, 4 D. 1, 0, -4 E. 0, -2, 8

i do not know the answer i am sorry

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.