Question

In the given problem, solve and graph.

Express answers in both inequality and interval notation.

$\frac{{(x+1)}^2}{x^2+2x-3}\le 0$

Answer 1

$\frac{{(x+1)}^2}{x^2+2x-3}\le 0$

$\frac{{(x+1)}^2}{(x-1)(x+3)}\le 0$

	x<-3	x=-3	-3<x<-1	x=-1	-1<x<1	x=1	x>1
x+1	-	-	-	0	+	+	+
x-1	-	-	-	-	-	0	+
x+3	-	0	+	+	+	+	+
$\frac{{(x+1)}^2}{(x-1)(x+3)}$	+	undefined	-	0	-	undefined	+

$\left[\begin{array}{cc}Solution:& -3<x<1\\ IntervalNotation:& (-3,1)\end{array}\right]$