Question

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

Answer 1

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

$(2-x^2)=(\sqrt{2}-x)(\sqrt{2}+x)\Rightarrow$ Critical points $\sqrt{2}$

${(x-3)}^2\Rightarrow$ Critical point $\left(3\right)$

$(x+1)\Rightarrow$ Critical point$(-1)$

$x^2-3x-4=0\Rightarrow x^2-4x+x-4=0$

or, $x(x-4)+1(x-4)=0$

or, $(x+1)(x-4)$

$\therefore$ From $1$

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

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

$\lambda ϵ(3,4)\cup (-1,\sqrt{2})\cup (-\sqrt{2},-1)$

$\lambda ϵ(-\sqrt{2},-1)\cup (-1,\sqrt{2})\cup (3,4)$