Question 38925
<pre><font size = 4><b>xｳ + 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): ｱ1, ｱ2

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

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

xｳ + 0xｲ + x - 2 = 0

1 | 1  0  1 -2
  |<u>    1  1  2</u> 
    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

xｳ + x + 2 = 0

as

(x - 1)(xｲ + x + 2) = 0

Us the 0-factor principle:

x - 1 = 0 gives solution x = 1

xｲ + x + 2 = 0 must be solved by the
quadratic formula:
              ____________
      -(1) ｱ <font face = "symbol">ﾖ</font>(1)ｲ-4(1)(2) 
x =  覧覧覧覧覧覧覧覧覧覧覧
              2(1)

            ______
      -1 ｱ <font face = "symbol">ﾖ</font>1 - 8 
x =  覧覧覧覧覧覧覧
            2

            __
      -1 ｱ <font face = "symbol">ﾖ</font>-7 
x =  覧覧覧覧覧
          2

             _
      -1 ｱ i<font face = "symbol">ﾖ</font>7 
x =  覧覧覧覧覧
          2
            _              
x = -1/2 ｱ <font face = "symbol">ﾖ</font>7/2ｷi

So the three solutions are
            _                  _ 
1,  -1/2 + <font face = "symbol">ﾖ</font>7/2ｷi, and -1/2 - <font face = "symbol">ﾖ</font>7/2ｷi 

Edwin
AnlytcPhil@aol.com</pre>