Question 1011575
One can guess, using possible roots of +/-1,+/-2, +/-3, +/-4, +/-6, and +/-12.
The latter two are unlikely.
try 
1:1//2//1//8//-12
=1=3=4=12=/0
(x-1) is a factor
What is left is x^3+3x^2+4x+12
-3:1//3//4//12
==1/0/4//0
(x+3) is a factor
left is x^2+4, which factors into (x+2i) and (x-2i)
The zeros are 1,-3, +2i,-2i
{{{graph(300,200,-10,10,-100,100,x^4+2x^3+x^2+8x-12)}}}