# How do you solve x(x-1)(x+2)>0?

Dec 7, 2016

The answer is x in ] -2,0 [ uu ] 1,+oo [

#### Explanation:

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

We do 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}$$- 2$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{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}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

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

$\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}$$+$

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

graph{x(x-1)(x+2) [-8.89, 8.89, -4.444, 4.445]}

Dec 7, 2016

The solution for the inequality is $- 2 < x < 0$ or $x > 1$.

#### Explanation:

The inequality given is $x \left(x - 1\right) \left(x + 2\right) > 0$ i.e. product of all the terms is positive. It is apparent that sign of terms $\left(x + 2\right)$, $x$ and $\left(x - 1\right)$ will change around the values $- 2$, $0$ and $1$ respectively. In sign chart we divide the real number line using these values, i.e. below $- 2$, between $- 2$ and $0$, between $0$ and $1$ and above $1$ and see how the sign of $x \left(x - 1\right) \left(x + 2\right)$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - 2 \textcolor{w h i t e}{X X X X X} 0 \textcolor{w h i t e}{X X X X X} 1$

$\left(x + 2\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} + i v e$

$x \textcolor{w h i t e}{X X X X X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} + i v e$

$\left(x - 1\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

$x \left(x - 1\right) \left(x + 2\right)$
$\textcolor{w h i t e}{X X X X X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

It is observed that $x \left(x - 1\right) \left(x + 2\right) > 0$ when either $- 2 < x < 0$ or $x > 1$, which is the solution for the inequality.