f(-1) = (1)^4+(-1)^3-13(-1)^2-(-1)+12 = 0

Therefore, x = -1

Dividing f(x) by x+1 the quotient would be x^3-13x^2+12

let f'(x) = x^3-13x^2+12

substituting x = 1 in f'(x) we get

f'(1) = 1 - 13 + 12 = 0

Therefore, x = 1 is also a zero of f(x)

The quotient would be x^2 + x - 12

on factorizing, (x+4)(x-3) = 0 which implies x = -4 and x = 3

Therefore, the solution to f(x) is x = -1 , 1 , -4 , 3   [ Answer]