The function is
.
Let
.
Perform the synthetic division method by testing
.

Since
, conclude that
is a zero of
.
Therefore,
is a rational zero.
The remaining factor can be written as
.


Therefore
has
real zeros at
,
and
.
Create a sign chart using these values.
\.gif\")
Note : The hallow circles denote that the values are not included in the solution set.
\Observe the sign chart :
\The set of values of
denoted in blue color represents the solution set.
Determine whether
is positive or negative on the test intervals.
Test intervals are
and
.
If
, then
.
If
, then
.
If
, then
.
If
,then
.
The solutions of
are
values such that
is positive.
is positive in the interval
and
.
The solution is
.
The solution set of
is
.