We factor the denominator



Factor out x2 from the first two terms and
factor out +1 from the last two terms:

 

Factor out (x-2)

 



Write 1 as 



The LCD is (x-2)(x²+1)

Multiply each numerator and denominator by whatever
factor(s) that are needed so the resulting denominator
will become the LCD.



Combine all the numerators over the LCD:






Combining terms:



That numerator x3+x-2 obviously has zero 1.
So we factor it using synthetic division:

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

So we see that it factors as (x-1)(x2+x+2)



The critical numbers are real zeros of the numerator and
zeros of the denominator.

The only real zero of the numerator is 1
The only real zero of the denominator is 2

We place those critical numbers on a number line
They cannot be solutions themselves because the 
inequality does not permit equality:

-----------o----o----------
-1    0    1    2    3    4

We choose a test value in the interval left of 1
The easiest one is 0 and substitute it into the
inequality:







That is true, so we shade the number line left of 1

<==========o----o----------
-1    0    1    2    3    4

We choose a test value between 1 and 2. The easiest 
one is 1.5 and substitute it into the
inequality:







That is false, so we do not shade the number line between
1 and 2.  So we still have this graph:

<==========o----o----------
-1    0    1    2    3    4

We choose a test value in the interval right of 2
The easiest one is 3 and substitute it into the
inequality:







That is true, so we shade the number line right of 2

<==========o----o==========>
-1    0    1    2    3    4

That is the graph of the solution.  In interval notation
that is written:



Edwin