What are the zeros of the polynomial function f(x)=x^3-4x^2-12x

a. 0,2,-6
b. -2,6
c. 0,-2,6
d. 2,-6

C: x = -2, 0, and 6

Step-by-step explanation:

First set the equation equal to zero

0 = x³ - 4x² - 12x

now factor out the GCF, which is 'x'

0 = x(x² - 4x - 12)

Now factor completely...

0 = x(x - 6)(x + 2)

So

x = 0

and

x - 6 = 0   becomes x = 6  (add 6 to both sides)

and

x + 2 = 0  becomes x = -2  (subtract 2 from both sides)

So our solutions are

x = -2, 0, and 6

c

step-by-step explanation:

idk

step-by-step explanation: