Question

Solve : $\frac{(x+1)(x-2{)}^2}{(x+3)(x^2+4x+10)}\ge 0$

Answer 1

$\frac{(x+1){(x-2)}^2}{(x+3)(x^2+4x+10)}\ge 0$

${(x-2)}^2$ is always $\ge 0$

$x^2+4x+10$ is always $>0$ ($\because$ $b^2<4ac\&a>0$)

$\therefore$ $\frac{(x+1)}{(x+3)}\ge 0$

$\therefore$ $x\ne -3$

also $x\ge -1$ or $x<-3$

$\therefore$ $xϵ(-\infty ,-3)\cup (-1,\infty )$