x³ + x - 2 = 0

The possible rational solutions have numerators which divide 
evenly into the absolute value of the last term.  The
possible denominators are those that divide evenly into
the absolute value of the coefficient of the highest power
of x. 

Possible numerators: divisors of 2 which are: 1, 2

Possible denoiminators: divisors of 1, of which there is only one: 1

Possible rational solutions (they can be positive or negative): ±1, ±2

Try x = 1, using synthetic division
 
x³ + x - 2 = 0

We must first insert a place-holder term for x², namely 0x²

x³ + 0x² + x - 2 = 0

1 | 1  0  1 -2
  |    1  1  2 
    1  1  2  0

The last number on the bottom is zero, so
we are lucky to have found a solution on
the first trial.

Now we have factored the left side of

x³ + x + 2 = 0

as

(x - 1)(x² + x + 2) = 0

Us the 0-factor principle:

x - 1 = 0 gives solution x = 1

x² + x + 2 = 0 must be solved by the
quadratic formula:
              ____________
      -(1) ± Ö(1)²-4(1)(2) 
x =  ——————————————————————
              2(1)

            ______
      -1 ± Ö1 - 8 
x =  ——————————————
            2

            __
      -1 ± Ö-7 
x =  ——————————
          2

             _
      -1 ± iÖ7 
x =  ——————————
          2
            _              
x = -1/2 ± Ö7/2·i

So the three solutions are
            _                  _ 
1,  -1/2 + Ö7/2·i, and -1/2 - Ö7/2·i 

Edwin
AnlytcPhil@aol.com