Answer
$\left(-\infty, -2\right) \cup (1, 2)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{5}{x^3-x^2-4x+4} \lt 0$
$\displaystyle \frac{5}{x^2(x-1)-4(x-1)}\lt0,$
$\displaystyle \frac{5}{(x^2-4)(x-1)}\lt0,$
$\displaystyle \frac{5}{(x-2)(x+2)(x-1)}\lt0,$
2. The cut points are:
$\displaystyle \frac{5}{(x-2)(x+2)(x-1)}= 0$
$x=-2$ or $x=2$ or $x=1$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \lt 0 ? \\
& & \displaystyle \frac{5}{(a-2)(a+2)(a-1)} & \\
(-\infty,-2) & -10 & \frac{}{(-)(-)(-)}=(-) & T\\
(-2,1) & 0 & \frac{}{(-)(+)(-)}=(+) & F\\
(1, 2) & \frac{3}{2} & \frac{}{(-)(+)(+)}=(-) & T\\
(2,\infty) & 10 & \frac{}{(+)(+)(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $\left(-\infty, -2\right) \cup (1, 2)$
