Question 331300
Factor the polynomial {{{x^3+8x^2+19x+12}}}

Since the highest degree is 3, there exist 3 roots to this poynomial

The real roots (there might be some complex roots) will be from the set 
 -/+1, -/+2, -/+3, -/+4,-/+6, -/+12

Use Discarte's sign rule
Number of positive roots P(x) ++++ no sign change, so no positive roots
number of negative roots P(-x) -+-+ 3 sign changes, so either 3 negative roots or 1 negative root and 2 complex roots

Since we know that at least one of the roots is negative, the roots are from 
the set  -1,-2,-3,-4,-6,-12
so experiment with -1 and see  {{{P(-1)=(-1)^3 + 8(-1)^2+19(-1) +12 = -1+8-19+12=0}}} so -1 is a root.

Now apply either long division or synthetic division to reduce the polynomial by one degree yield  {{{x^2+7x+12}}} this can be easily factored into (x+3)(x+4)

so the original polymial can be factord into 
{{{x^3+8x^2+19x+12 = (x+1)*(x+3)*(x+4)}}}