SOLUTION: factor the polynomial completely x^3+8X^2+19X+12

All feasible rational zeros are  the factors of 12 


 , , , , , ,   

Also there are no sign changes so there are no positive zeros.

So we begin by trying -1


 -1| 1  8 19  12
   |   -1 -7 -12
     1  7 12   0

So x = -1 is a zero, so therefore (x + 1) is a factor.
So now we have factored the original polynomial as





Now we factor the trinomial on the right and get:



That's it.

Edwin