Question

Solve the following inequality:
$\frac{(x+2)(x^2-2x+1)}{4+3x-x^2}⩾0$

Answer 1

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

$=>\frac{(x+2)(x^2-2x+1)}{(x^2-3x-4)}\le 0$

$=>\frac{(x+2){(x-1)}^2}{(x^2-4x+x-4)}\le 0$

$=>\frac{(x+2){(x-1)}^2}{x(x-4)+1(x-4)}\le 0$

$=>\frac{(x+2){(x-1)}^2}{(x+1)(x-4)}\le 0$

Hence, $xϵ(-\infty ,-2]\cup [-1,1)\cup (1,4)$.