I need help finding the zeros of a function algebraically. And here is one of them f(x)= x^3-4x^2-9x+36

Answer 1

try writing as:
(x^3 - 9x) -(4x^2-36) =0

x(x^2-9) -4 (x^2-9)=0

(x^2-9)(x-4) = 0

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

3 , -3 , 4