f(x)=x^3+ax^2+bx+12

f(x) is divisible by (x+1)
∴ f(-1)=0
f(-1)=(-1)^3+a(-1)^2+b(-1)+12=0
-1+a-b+12=0
a-b+11=0 . . . . . (1)

It leaves a remainder 6 when it is divided by (x-2)
∴ f(2)=6
f(2)=2^3+a(2)^2+2b+12=6
8+4a+2b+12=6
2a+b+7=0 . . . . . (2)

(1)+(2) : 3a+18=0
a=-6
Substitute a=-6 into (1) :
-6-b+11=0
b=5

∴ f(x)=x^3-6x^2+5x+12
=(x+1)(x^2-7x+12)
=(x+1)(x-3)(x-4)

f(x)=0
(x+1)(x-3)(x-4)=0
x+1=0 or x-3=0 or x-4=0
∴ x=-1 or x=3 or x=4