# How do you solve x^3+7x^2-x-7<0?

Jan 29, 2017

The answer is x in ]-oo, -7 [uu]-1, 1[

#### Explanation:

Let $f \left(x\right) = {x}^{3} + 7 {x}^{2} - x - 7$

$f \left(1\right) = 1 + 7 - 1 - 7 = 0$

so, $\left(x - 1\right)$ is a factor of $f \left(x\right)$

To find the other factors, we do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} + 7 {x}^{2} - x - 7$$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{b l u e}{x - 1}$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$|$$\textcolor{red}{{x}^{2} + 8 x + 7}$

$\textcolor{w h i t e}{a a a a a}$$0 + 8 {x}^{2} - x$

$\textcolor{w h i t e}{a a a a a a a}$$+ 8 {x}^{2} - 8 x$

$\textcolor{w h i t e}{a a a a a a a a a}$$+ 0 + 7 x - 7$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$+ 7 x - 7$

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

Therefore,

$f \left(x\right) = \left(x - 1\right) \left({x}^{2} + 8 x + 7\right) = \left(x - 1\right) \left(x + 1\right) \left(x + 7\right)$

Now, we can build 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}$$- 7$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 7$$\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}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\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}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\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}$$+$

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

Therefore,

$f \left(x\right) < 0$ when x in ]-oo, -7 [uu]-1, 1[