Question 1090441
<pre><b>
{{{1-(2x^"")/(x^2+1)-(1+4x-3x^2)/(x^3-2x^2+x-2)+1^""/(x^""-2) >0}}}

We factor the denominator

{{{x^3-2x^2+x-2}}}

Factor out x<sup>2</sup> from the first two terms and
factor out +1 from the last two terms:

{{{x^2(x^""-2)+1(x-2)}}} 

Factor out (x-2)

 {{{(x^""-2)(x^2+1)}}}

{{{1-(2x^"")/(x^2+1)-(1+4x-3x^2)/((x^""-2)(x^2+1))+1^""/(x^""-2) >0}}}

Write 1 as {{{1/1}}}

{{{1^""/1^""-(2x^"")/(x^2+1)-(1+4x-3x^2)/((x^""-2)(x^2+1))+1^""/(x^""-2) >0}}}

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.

{{{(1(x^""-2)(x^2+1))/(1(x^""-2)(x^2+1))-((2x^"")(x^""-2))/((x^2+1)(x^""-2))-(1+4x-3x^2)/((x^""-2)(x^2+1))+(1(x^2+1))/((x^""-2)(x^2+1)) >0}}}

Combine all the numerators over the LCD:

{{{(1(x^""-2)(x^2+1)-(2x^"")(x^""-2)-(1+4x-3x^2)+1(x^2+1))/((x^""-2)(x^2+1)) >0}}}


{{{(x^3+x-2x^2-2-2x^2+4x-1-4x+3x^2+x^2+1)/((x^""-2)(x^2+1)) >0}}}

Combining terms:

{{{(x^3+x-2)/((x^""-2)(x^2+1)) >0}}}

That numerator x<sup>3</sup>+x-2 obviously has zero 1.
So we factor it using synthetic division:

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

So we see that it factors as (x-1)(x<sup>2</sup>+x+2)

{{{((x^""-1)(x^2+x+2))/((x^""-2)(x^2+1)) >0}}}

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:

{{{((0^""-1)(0^2+0+2))/((0^""-2)(0^2+1)) >0}}}

{{{((-1)(2))/((-2)(1))>0}}}

{{{1>0}}}

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:

{{{((1.5^""-1)(1.5^2+1.5+2))/((1.5^""-2)(1.5^2+1)) >0}}}

{{{((0.5)(5.75))/((-0.5)(3.25))>0}}}

{{{"-1.769...">0}}}

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:

{{{((3^""-1)(3^2+3+2))/((3^""-2)(3^2+1)) >0}}}

{{{((2)(14))/((1)(10))>0}}}

{{{2.8>0}}}

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:

{{{matrix(1,3,
(matrix(1,3,-infinity,",",1)),
"U",
(matrix(1,3,2,",",infinity)))}}}

Edwin</pre>