# How do you solve x^2+6x-16>0 using a sign chart?

Dec 24, 2016

The answer is x in ] -oo,-8 [ uu ]2, +oo[

#### Explanation:

Let's factorise the expression

${x}^{2} + 6 x - 16 = \left(x + 8\right) \left(x - 2\right)$

and let $f \left(x\right) = \left(x + 8\right) \left(x - 2\right)$

Now we can do the 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}$$- 8$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{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}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a 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}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) > 0$ when x in ] -oo,-8 [ uu ]2, +oo[