# How do you solve and write the following in interval notation:  x(x − 1)(x + 3) >0?

Oct 8, 2016

Solution in interval notation is $\left(- 3 , 0\right)$ or $\left(1 , \infty\right)$

#### Explanation:

We are solving the inequality $x \left(x - 1\right) \left(x + 3\right) > 0$. From this we know that the product $x \left(x - 1\right) \left(x + 3\right)$ is positive. It is also apparent that sign of binomials $\left(x + 3\right)$, $x$ and $\left(x - 1\right)$ will change around the values $- 3$. $9$ and $1$ respectively. We divide the real number line using around these values, i.e. below $- 3$, between $- 3$ and $0$, between $0$ and $1$ and above $1$ and see how the sign of  changes, as we move across he real number line.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - 3 \textcolor{w h i t e}{X X X X X} 0 \textcolor{w h i t e}{X X X X X} 1$

$\left(x + 3\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} + i v e$

$x \textcolor{w h i t e}{X X X X X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} + i v e$

$\left(x - 1\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

$x \left(x - 1\right) \left(x + 3\right)$
$\textcolor{w h i t e}{X X X X X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

It is observed that $x \left(x - 1\right) \left(x + 3\right) > 0$ when either $- 3 < x < 0$ or $x > 1$, which is the solution for the inequality.

This can be written in interval notation as $\left(- 3 , 0\right)$ or $\left(1 , \infty\right)$.

Here small brackets $\left(\right)$ indicate that endpoints are not included though solution includes values in the interval. This is used when we have only inequality..

If we have equality included say for $x \left(x - 1\right) \left(x + 3\right) \ge 0$, we use square brackets, which means endpoints are included.