how do you find all the real zeros for this function f(x)=x^3+9x^2-4x-36
your expression factors by grouping
let
x^3 + 9x^2 - 4x - 36 = 0
x^2(x+9) - 4(x+9) = 0
(x+9)(x^2-4) = 0
(x+9)(x+2)(x-2) = 0
x = -9, -2, 2
let
x^3 + 9x^2 - 4x - 36 = 0
x^2(x+9) - 4(x+9) = 0
(x+9)(x^2-4) = 0
(x+9)(x+2)(x-2) = 0
x = -9, -2, 2