# How do you solve x^{3} + 5x^{2} - x \geq 5?

Sep 22, 2016

either $- 5 \le x \le - 1$ or $x \ge 1$

#### Explanation:

${x}^{3} + 5 {x}^{2} - x \ge 5$ is equivalent to ${x}^{3} + 5 {x}^{2} - x - 5 \ge 0$

${x}^{3} + 5 {x}^{2} - x - 5 = x \left({x}^{2} - 1\right) + 5 \left({x}^{2} - 1\right) = \left(x + 5\right) \left({x}^{2} - 1\right) = \left(x + 5\right) \left(x + 1\right) \left(x - 1\right)$

Hence, we have $\left(x + 5\right) \left(x + 1\right) \left(x - 1\right) \ge 0$

From this we know that the product $\left(x + 5\right) \left(x + 1\right) \left(x - 1\right) \ge 0$ is positive or zero. It is apparent that sign of binomials $\left(x + 5\right)$, $\left(x + 1\right)$ and $\left(x - 1\right)$ will change around the values $- 5$. $- 1$ and $1$ respectively.

In sign chart we divide the real number line in four parts, i.e. below $- 5$, between $- 5$ and $- 1$, between $- 1$ and $1$ and above $1$ and see how the sign of $\left(x + 5\right) \left(x - 1\right) \left(x + 1\right)$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - 5 \textcolor{w h i t e}{X X X X X} - 1 \textcolor{w h i t e}{X X X X X} 1$

$\left(x + 5\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X X} + i v e$

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

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

$\left(x + 5\right) \left(x + 1\right) \left(x - 1\right)$
$\textcolor{w h i t e}{X X X X X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

It is observed that $\left(x + 5\right) \left(x + 1\right) \left(x - 1\right) \ge 0$ when either $- 5 \le x \le - 1$ or $x \ge 1$, which is the solution for the inequality.