Question

The set of the solutions for $\frac{(x+1)(x-3)}{(x+5)}<0$ is

• A. $(-\infty ,-5)\cup (-1,3)$
• B. $(-\infty ,-5)\cup (-1,\infty )$
• C. $(-5,-1)\cup (3,\infty )$
• D. $(-\infty ,-1)\cup (3,\infty )$

Answer 1

The correct option is A $(-\infty ,-5)\cup (-1,3)$

$\frac{(x+1)(x-3)}{(x+5)}<0$


$\therefore x\in (-\infty ,-5)\cup (-1,3)$