#### Answer

$$(-∞, -2) ∪ (4,+∞)$$

#### Work Step by Step

Let $(x-4)(x+2)>0$ be a function $f$.
Find the $x$-intercepts by solving $(x-4)(x+2)=0$
$x-4=0$ or $x+2=0$
$x = 4$ or $x = -2$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, -2)(-2,4)(4, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -2)$
Test Value = $-3$
$f=(x-4)(x+2)$
$f=(-3-4)(-3+2)$
$f=7$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -2)$.
$(-2,4)$
Test Value = $0$
$f=(x-4)(x+2)$
$f=(0-4)(0+2)$
$f=-8$
Conclusion: $f (x) < 0$ for all $x$ in $(-2,4)$.
$(4,+∞)$
Test Value = $5$
$f=(x-4)(x+2)$
$f=(5-4)(5+2)$
$f=7$
Conclusion: $f (x) > 0$ for all $x$ in $(4,+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given
inequality. We are interested in solving $f (x) > 0$, where $f (x) = (x-4)(x+2)$. Based on the solution above, $f(x)>0$ for all $x$ in $(-∞, -2)$ or $(4,+∞)$.
Thus, the solution set of the given inequality is:
$$(-∞, -2) ∪ (4,+∞)$$