# How do you solve x^3+2x^2-x-2>=0?

##### 1 Answer
Jan 17, 2017

The answer is x in [-2,-1] uu [1,+ oo[

#### Explanation:

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

$f \left(1\right) = 1 + 2 - 1 - 2 = 0$

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}$${x}^{3} + 2 {x}^{2} - x - 2$$\textcolor{w h i t e}{a a a a}$∣$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}$∣${x}^{2} + 3 x + 2$

$\textcolor{w h i t e}{a a a a}$$0 + 3 {x}^{2} - x$

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

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

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

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

Therefore,

$\frac{{x}^{3} + 2 {x}^{2} - x - 2}{x - 1} = {x}^{2} + 3 x + 2 = \left(x + 1\right) \left(x + 2\right)$

So,

$f \left(x\right) = \left(x + 2\right) \left(x + 1\right) \left(x - 1\right)$

Now, we can construct the sign chart

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

$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a 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}$$+$$\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 a}$$-$$\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}$$x - 1$$\textcolor{w h i t e}{a a 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}$$-$$\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 a}$$-$$\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}$$+$

Therefore,

$f \left(x\right) \ge 0$, when x in [-2,-1] uu [1,+ oo[