# How do you solve (x-3)^2(2x+1)<=0 using a sign chart?

May 31, 2017

The solution is $x \in \left(- \infty , - \frac{1}{2}\right] \cup \left\{3\right\}$

#### Explanation:

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

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$

$\forall x \in \mathbb{R}$, ${\left(x - 3\right)}^{2} \ge 0$

The sign chart is very simple

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \frac{1}{2}$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$2 x + 1$$\textcolor{w h i t e}{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}$${\left(x - 3\right)}^{2}$$\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}$$f \left(x\right)$$\textcolor{w h i t e}{a 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}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left(- \infty , - \frac{1}{2}\right] \cup \left\{3\right\}$