x5-3x4-8x3-8x2-9x-5=0

The possible rational zeros are ± the factors of 5, 
so they are ±1, and ±5

We try the easiest one x=1

1|1 -3  -8  -8  -9  -5
 |   1  -2 -10 -18 -27 
  1 -2 -10 -18 -27 -32

No that gives remaider -32, not 0, so 1 is not a solution

We try the next easiest one x=-1

-1|1 -3  -8 -8 -9 -5
  |  -1   4  4  4  5 
   1 -4  -4 -4 -5  0

That gives remaider 0, so -1 is a solution and so we have
factored the polynomial on the left as:

(x + 1)(x4 - 4x³ - 4x² - 4x - 5) = 0 

So now we try to factor that polynomial in the second
parentheses:

It also has the same set of possible solutions, so we try -1
again (there is no use to try 1) 

-1|1 -4 -4 -4 -5 
  |  -1  5 -1  5 
   1 -5  1 -5  0

That gives remaider 0, so -1 is a DOUBLE solution 
and so we have further factored the polynomial on the left:

(x + 1)(x + 1)(x³ - 5x² + x - 5) = 0

We can now factor the polynomial in the third parentheses by
grouping:

               x³ - 5x² + x - 5
              x²(x - 5) + 1(x - 5)
              (x - 5)(x² + 1)

So we have now factored the original left side as

(x + 1)(x + 1)(x - 5)(x² + 1) = 0

Now we use the zero-factor principle and set each factor = 0

x + 1 =  0,  x + 1 = 0,  x - 5 = 0, x² + 1 = 0
    x = -1       x = -1      x = 5.     x² = -1
                                         x = 
                                         x = ±i

So the solutions are  -1, -1, 5, i, -i.

-1 was a solution twice, so we call it a "double root".

Edwin