Question

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

Answer 1

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

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

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

$=>xϵ[-\sqrt{2},-1)\cup (1,\sqrt{2}]\cup [3,4)$.