Solve x^5 + 7x^3 + 6x

`x^5 + 7x^3 + 6x < 5x^4 + 7x^2 +
2`

`x^5 -5x^4 + 7x^3- 7x^2 + 6x - 2 <
0`

 Let us solve the
equation:

`x^5 - 5x^4 + 7x^3 - 7x^2 + 6x -2 =
0`

We will substitute with x=
1

==> 1 - 5 + 7 - 7 + 6 -2 =
0

==> 2 - 2 =
0

==> 0 = 0

Then x= 1
is a root for the equation.

==> (x-1) is a
factor:

`x^5 -5x^4 +7x^3 -7x^2 + 6x -2= (x-1)(x^4 -4x^3
+3x^2-4x +2)`

Now we will factor the last
terms.

First we will rearrange
terms.

`x^4 - 4x^3 + 3x^2 - 4x +2 = (x^4+3x^2 +2) +
(-4x^3-4x)`

`==> (x^4+3x^2 + 2)=
(x^2+1)(x^2+2)`

`==> (x^2+1)(x^2+2)
-4x(x^2+1)`

Now we will factor
x^2+1

`==> (x^2+1)(x^2+2 - 4x) = (x^2
+1)(x^2-4x+2)`

Now we will rewrite the
inequality:

`==> (x-1)(x^2+1)(x^2-4x+2) <
0`

`` We know that `x^2+1 `  is always greater than
zero.

`==> (x-1)(x^2-4x+2) <
0`

`` Now we will find the
roots.

==> x1=
1

==> `x2= (4+sqrt(16-8))/ 2 = (4+sqrt(8))/2 =
2+sqrt2`

`==> x3= 2-
sqrt2`

`` ==> Now we have the
intervals:

`(-oo, 1), (1, 2-sqrt2) , (2-sqrt2 , 2+sqrt2) ,
and (2+sqrt2, oo)`

`( -oo, 1) ==>
+`

`(1, 2-sqrt2) ==>
-`

`(2,2-sqrt2, 2+sqrt2) ==>
+`

`(2+sqrt2, oo) ==>
-`

Then, solution is:

x `in (
1, 2-sqrt2) U (2+sqrt2,
oo)`

``