# How do you solve x^3+7x^2+10x>=0?

Oct 1, 2016

Solution is $- 5 \ge x \ge - 2$ or $x \ge 0$

#### Explanation:

Let us first factorize ${x}^{3} + 7 {x}^{2} + 10 x$.

${x}^{3} + 7 {x}^{2} + 10 x = x \left({x}^{2} + 7 x + 10\right) = x \left({x}^{2} + 2 x + 5 x + 10\right)$

= x(x(x+2)+5(x+2)=x(x+2)(x+5)

Hence we have to solve the inequality ${x}^{3} + 7 {x}^{2} + 10 x > 0$ or $\left(x + 5\right) \left(x + 2\right) x \ge 0$

From this we know that for the product $\left(x + 5\right) \left(x + 2\right) x \ge 0$, signs of binomials $\left(x + 5\right)$, $\left(x + 2\right)$ and $x$ will change around the values $- 5$. $- 2$ and $0$ respectively. In sign chart we divide the real number line around these values, i.e. below $- 5$, between $- 5$ and $- 2$, between $- 2$ and $0$ and above $0$ and see how the sign of $\left(x + 5\right) \left(x + 2\right) x$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - 5 \textcolor{w h i t e}{X X X X X} - 2 \textcolor{w h i t e}{X X X X X} 0$

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

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

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

$\left(x + 5\right) \left(x + 2\right) x$
$\textcolor{w h i t e}{X X X X X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

It is observed that $\left(x + 5\right) \left(x + 2\right) x \ge 0$ when either $- 5 \ge x \ge - 2$ or $x \ge 0$, which is the solution for the inequality.