Horner's method is simply synthetic division.

x³ + 5x² + 8x + 6 = 0

The feasible rational roots are ± the factors of 6, which are

±1, ±2, ±3, ±6

It has no sign changes (all terms positive), so there are
no positive roots.  So we need try only negative feasible
rational roots:

We try -1

-1|1  5  8  6 
  |  -1 -4 -4 
   1  4  4  2

No, that doesn't give 0 on the far right.

We try -2

-2|1  5  8  6 
  |  -2 -6 -4 
   1  3  2  2

No, that doesn't give 0 on the far right, either.

We try -3

-3|1  5  8  6 
  |  -3 -6 -6 
   1  2  2  0

Whoopie do! That give 0 on the far right!

So we have have found one zero -3. The above
synthetic division is a shortcut for long division
of dividing by (x + 3) and getting x² + 2x + 2 as a
quotient, so we have now factored:

   x³ + 5x² + 8x + 6 = 0

as

(x + 3)(x² + 2x + 2) = 0

Using the zero factor principle:

x + 3 = 0;       x² + 2x + 2 = 0
    x = -3       x = 
                 x =              
                 x = 
                 x = 
                 x = 
                 x = 
                 x = 
                 x = -1 ± i

So the three roots are -3, -1 + i, -1 - i

Edwin