How do you solve $\frac{(x + 3) (x - 2)}{x + 1} < 0$ $((x+3)(x-2))/(x+1)<0$ ?

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Answer 1 · 2018-07-20T19:57:31

$x \in (- \infty, - 3) \cup (- 1, 2)$ $x\in(-\infty, -3)\cup(-1, 2)$

Given inequality

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

Critical points of above inequality are

$x = 2, - 1, - 3$ $x=2, -1, -3$

Taking all the above critical points on the number line & dividing the regions in positive & negative intervals, we get the solution

$x \in (- \infty, - 3) \cup (- 1, 2)$ $x\in(-\infty, -3)\cup(-1, 2)$

How do you solve ((x+3)(x-2))/(x+1)<0(x+3)(x−2)x+1<0((x+3)(x-2))/(x+1)<0?