Question 970668
The possible factors are +/- 1  +/-2  and +/-4

Let's try (x-1)

1 ;;   1 ;; 1  ;2 ;;-4
============1 ==2  ;=4 

======  1===2==4==0

(x-1) (x^2+2x+4)=0

One root is 1; the second does not factor (4ac is larger than b^2=see below)

The quadratic formula for the other 2
=[-2 +/- sqrt (4-16)]/2  ;;   sqrt (-12)=sqrt (3)* sqrt (-4)= 2i (sqrt (3)

-1 +/- i* sqrt (3)

{{{graph (300,300,-10,10,-10,10,x^3+x^2+2x-4)}}}

See a graph of a cubic equation with a root at x=1 and the y-intercept at (0.-4) consistent with the equation