Question

Solve the inequality. Express the solution using interval notation and graph the solution set on the real number line.

$\frac{5}{x^3-x^2-4x+4}<0$

Short answer

The solution of the inequality is$\left(-\infty ,-2\right)\cup \left(1,2\right)$ and the graph is shown in figure (1).

Step-by-step solution

Step 1. Write the given inequality.

The given inequality is,

$\frac{5}{x^3-x^2-4x+4}<0$

The above inequality can be simplified as,

$\begin{array}{c}\frac{5}{x^3-x^2-4x+4}<0\\ \frac{5}{x^2\left(x-1\right)-4\left(x-1\right)}<0\\ \frac{5}{\left(x^2-4\right)\left(x-1\right)}<0\\ \frac{5}{\left(x+2\right)\left(x-2\right)\left(x-1\right)}<0\end{array}$

The factors of the left hand side are$\left(x-1\right),\left(x-2\right)$ and $\left(x+2\right)$. These are zero when x is$1,2$ and $-2$. These numbers divide the real into the intervals $\left(-\infty ,-2\right),\left(-2,1\right),\left(1,2\right),\left(2,\infty \right)$

Step 2. Draw a table for test point.

Draw the table to determine the sign of each factor in each interval using test points.

	$\left(-\infty ,-2\right)$	$\left(-2,1\right)$	$\left(1,2\right)$	$\left(2,\infty \right)$
Sign of $\left(x+2\right)$	$-$	$+$	$+$	$+$
Sign of $\left(x-2\right)$	$-$	$-$	$-$	$+$
Sign of $\left(x-1\right)$	$-$	$-$	$+$	$+$
Sign of $\left(x-1\right)\left(x+2\right)\left(x-2\right)$	$-$	$+$	$-$	$+$
Sign of $\frac{5}{\left(x-1\right)\left(x+2\right)\left(x-2\right)}$	$-$	$+$	$-$	$+$

From the above table it is observed that equation consist of all numbers less than $-2$, and between 1 and 2.

So, the solution of the inequality in interval notation is $\left(-\infty ,-2\right)\cup \left(1,2\right)$.

Step 3. Plot the graph.

The solution of the inequality in interval notation is $\left(-\infty ,-2\right)\cup \left(1,2\right)$.

Plot the graph of the above interval on the real number line.

Figure (1)

Thus, the solution of the inequality is$\left(-\infty ,-2\right)\cup \left(1,2\right)$ and the graph is shown in figure (1).