Question

PLEASE HELP ASAP!!!

(k12 students appreciated!!)

What are the zeros of the function?
ƒ(x)= x^3+x^2-6x

Answer 1

Answer: x = 0, x = -3, x = 2

Explanation:

f(x) = x^3 + x^2 - 6x

f(x) = 0

x^3 + x^2 - 6x = 0

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

x = 0 or

x^2 + x - 6 = 0

D = b^2 - 4ac = 1 - 4*(-6) = 25

x = (-b - sqrt(D))/(2a) = (-1 - 5)/2 = -3 or

x = (-b + sqrt(D))/(2a) = (-1 + 5)/2 = 2