Simplify the expression $f\left(x\right)=\left(x^2-8\right)\left(x^2+6\right)$

$\left(2x+5\right)\left(2x-3\right)\ge3x-12$

$4\left(x-1\right)\left(x-5\right)>0$

$\left(x-2\right)\left(x+3\right)\ge0$

$\left(x+2\right)\cdot\left(x-1\right)<16$

$\left(x+2\right)\left(x-6\right)<8$

$\left(6+\sqrt{7}\right)\left(5+\sqrt{7}\right)$

$\left(x-4\right)\left(x+5\right)<\left(x-3\right)\left(x-2\right)$