Question

The number of integral value(s) of $x$ that satisfying the inequality $\frac{(2-x^2{)}^3(x-3{)}^5}{(x+1)(x^2-3x-4)}\ge 0$ is

Answer 1

$\frac{(2-x^2{)}^3(x-3{)}^5}{(x+1)(x^2-3x-4)}\ge 0$
$\Rightarrow \frac{(x^2-2{)}^3(x-3{)}^5}{(x+1)(x+1)(x-4)}\le 0$
$\Rightarrow \frac{(x+\sqrt{2}{)}^3(x-\sqrt{2}{)}^3(x-3{)}^5}{(x+1{)}^2(x-4)}\le 0$
$\Rightarrow x\in [-\sqrt{2},-1)\cup (-1,\sqrt{2}]\cup [3,4)$
Integers are $0,1,3$