I need help finding all imaginary and real roots?

f (-3) = - 81 + 36 + 81 - 36 = 0
Thus x + 3 is a factor of f (x)

x + 3 = x - (3)

To find other factors use synthetic division :-

- 3 | 3_____4_____- 27_____- 36
__ |______- 9_____ 15_____ 36
__ | 3____ - 5_____- 12_____ 0

(x + 3) (3x² - 5x - 12) = 0

(x + 3)(3x + 4)(x - 3) = 0

x = - 3 , x = - 4/3 , x = 3

3x^3 + 4x^2 - 27x - 36 = 0
(3x^3 + 4x^2) - (27x + 36) = 0
x^2(3x + 4) - 9(3x + 4) = 0
(3x + 4)(x^2 - 9) = 0
(3x + 4)(x^2 - 3^2) = 0
(3x + 4)(x + 3)(x - 3) = 0

3x + 4 = 0
3x = -4
x = -4/3

x + 3 = 0
x = -3

x - 3 = 0
x = 3

∴ x = -4/3 , ±3

3x^3+4x^2-27x-36=0=>
(3x^3-27x)+(4x^2-36)=0=>
3x(x^2-9)+4(x^2-9)=0=>
(x^2-9)(3x+4)=0=>
x=+/-3 or x=-4/3
All rts are real.

But where is the equation? You has given a polynomial only.

Or you meant 3x^3+4x^2-27x-36 = 0? Be careful. Somebody's time goes waste.
Take first and third term and second and fourth.
So 3x( x^2 - 9) +4(x^2 - 9) = 0.
(x^2 - 9)(3x+4)=0
so x = 3 or - 3 or - 4/3