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