$\frac{-24x^8}{8x^2}$

$4\left(1+m\right)^2-4\left(1+m\right)\left(n-1\right)+\left(n-1\right)^2$

$f\left(x\right)=x^2\left(1+x\right)^2$

$x ^ { 2 } - 8 x + 11 = 0$

$\lim_{x\to1}\left(\frac{1}{\ln\left(x\right)}-\frac{1}{x^2-1}\right)$

$0=18x^5+30x^4+12x^2+30x^3$

$\left(x+11\right)\left(x+9\right)$