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

Dec 16, 2016

The answer is x in ] -3,1 [

#### Explanation:

Let's factorise the denominator

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

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

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 3 , 1\right\}$

The numerator ${\left(x + 1\right)}^{2} > 0 , \forall x \in {D}_{f} \left(x\right)$

Now we can do the 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}$$- 3$$\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}$$x + 3$$\textcolor{w h i t e}{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}$$x - 1$$\textcolor{w h i t e}{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}$$f \left(x\right)$$\textcolor{w h i t e}{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}$$+$

Therefore,

$f \left(x\right) \le 0$ when x in ] -3,1 [

graph{y-(x+1)^2/(x^2+2x-3)=0 [-12.66, 12.65, -6.33, 6.33]}