All the possible rational roots have numerators which are divisors
of 10, positive or negative, and whose denominators are the divisors of
the coefficient of x³, which is 1.  Since 1 has only 1 divisor, 1, the
only possible roots are ± the divisors of 10.

±1, ±2, ±5, ±10

Try 1

1|1 -2 -17  10
 |   1  -1 -18
  1 -1 -18  -8

No, 1 is not a root because we get -8 
on the bottom right as a remainder, not 0.

-1|1 -2 -17  10
  |  -1   3  14
   1 -3 -14  24

No, -1 is not a root because we get 24 
on the bottom right as a remainder, not 0.

2|1 -2 -17  10
 |   2   0 -34
  1  0 -17 -24

No, 2 is not a root because we get -24 
on the bottom right as a remainder, not 0.

-2|1 -2 -17  10
  |  -2   8  18
   1 -4  -9  28

No, -2 is not a root because we get 28 
on the bottom right as a remainder, not 0.

 5|1 -2 -17  10
  |   5  15 -10
   1  3  -2   0

Yes!! Finally!! 5 is a root because we get 0 
on the bottom right.

So we have now factored the polynomial 

f(x) = x³ - 2x² - 17x + 10

as

f(x) = (x - 5)(x² + 3x - 2)

The trinomial in the parentheses does not factor,
so we set each factor = 0

Setting the first factor = 0:

    x - 5 = 0

        x = 5

Setting the second factor, the trinomial, = 0:

   x² + 3x - 2 = 0

We use the quadratic formula:

a = 1,  b = 3, c = -2



  







So the three roots are

5, , 

Edwin