Answer
See below:
Work Step by Step
The boundary points of the given inequality $\left( x-4 \right)\left( x+2 \right)>0$ , will be calculated by equating $f\left( x \right)$ to 0 as $\left( x-4 \right)\left( x+2 \right)=0$.
Then,
$x=4$ and $x=-2$
Now, draw the number line.
Now, $x=4$ and $x=-2$ divides the number line in three intervals.
$\left( -\infty ,-2 \right),\left( -2,4 \right),\left( 4,\infty \right)$
Now, testing the condition for $\left( x-4 \right)\left( x+2 \right)>0$ in the interval $\left( -\infty ,-2 \right)$.
Take the test point as $-3$ ,
$\begin{matrix}
\left( x-4 \right)\left( x+2 \right)>0 \\
\left( -3-4 \right)\left( -3+2 \right)\overset{?}{\mathop{>}}\,0 \\
\left( -7 \right)\left( -1 \right)\overset{?}{\mathop{>}}\,0 \\
7>0 \\
\end{matrix}$
As seen above, the condition is satisfied.
Now, testing the condition for $\left( x-4 \right)\left( x+2 \right)>0$ in the interval $\left( -2,4 \right)$, take the test point as $3$:
$\begin{matrix}
\left( x-4 \right)\left( x+2 \right)>0 \\
\left( 3-4 \right)\left( 3+2 \right)\overset{?}{\mathop{>}}\,0 \\
\left( -1 \right)\left( 5 \right)\overset{?}{\mathop{>}}\,0 \\
-5\overset{?}{\mathop{>}}\,0 \\
\end{matrix}$
As seen above, the condition is unsatisfied.
Now, testing the condition for $\left( x-4 \right)\left( x+2 \right)>0$ in the interval $\left( 4,\infty \right)$, take the test point as $5$ ,
$\begin{matrix}
\left( x-4 \right)\left( x+2 \right)>0 \\
\left( 5-4 \right)\left( 5+2 \right)\overset{?}{\mathop{>}}\,0 \\
\left( 1 \right)\left( 7 \right)\overset{?}{\mathop{>}}\,0 \\
7>0 \\
\end{matrix}$
As seen above, the condition is satisfied.
Therefore, the interval of the given inequality is $\left( -\infty ,-2 \right)\cup \left( 4,\infty \right)$.
