# How do you solve and write the following in interval notation: x^2 -3x -10 >0?

Jul 14, 2017

$\left(- \infty , - 2\right) \cup \left(5 , + \infty\right)$

#### Explanation:

$\text{factorise the left side}$

$\Rightarrow \left(x - 5\right) \left(x + 2\right) > 0$

$\text{the zeros are " x=5" and } x = - 2$

$\text{these indicate where the function may change sign}$
$\text{and 'splits the x-axis into 3 intervals, namely}$

$x < - 2 , \textcolor{w h i t e}{x} - 2 < x < 5 , \textcolor{w h i t e}{x} x > 5$

$\text{consider a "color(blue)" test point ""in each interval}$

$\text{substitute each test point into the function and consider }$
$\text{it's sign}$

$x = - 5 \to \left(-\right) \left(-\right) \to \textcolor{red}{\text{ positive}}$

$x = 1 \to \left(-\right) \left(+\right) \to \textcolor{b l u e}{\text{ negative}}$

$x = 6 \to \left(+\right) \left(+\right) \to \textcolor{red}{\text{ positive}}$

$\text{we want to find where the function is positive } > 0$

$\text{this occurs when "x<-2" or } x > 5$

$\Rightarrow \left(- \infty , - 2\right) \cup \left(5 , + \infty\right) \leftarrow \text{ interval notation}$
graph{x^2-3x-10 [-10, 10, -5, 5]}