# How do you solve and write the following in interval notation: (x + 6)(x + 8) ≥ 0?

Nov 24, 2017

The solution is $x \in \left(- \infty , - 8\right] \cup \left[- 6 , + \infty\right)$

#### Explanation:

Let $f \left(x\right) = \left(x + 6\right) \left(x + 8\right)$

Build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 8$$\textcolor{w h i t e}{a a a a a a a}$$- 6$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 8$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 6$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 8\right] \cup \left[- 6 , + \infty\right)$

graph{(x+6)(x+8) [-11.54, 0.947, -2.155, 4.09]}

Nov 24, 2017

$\left(- \infty , - 8\right] \cup \left[- 6 , \infty\right)$

#### Explanation: $\left(x + 6\right)$ is ZERO for $x = \left(- 6\right)$

$\left(x + 6\right)$ is POSITIVE for $x > \left(- 6\right)$

$\left(x + 8\right)$ is ZERO for $x = \left(- 8\right)$

$\left(x + 8\right)$ is POSITIVE for $x > \left(- 8\right)$

Refer to the SIGN Table

Our required Interval Notation can be written as follows:

$\left(- \infty , - 8\right] \cup \left[- 6 , \infty\right)$