Find the result of $44\left({\mathrm{x}}^4-5{\mathrm{x}}^3-24{\mathrm{x}}^2\right)÷11\mathrm{x}(\mathrm{x}-8)$.

$44\left({\mathrm{x}}^4-5{\mathrm{x}}^3-24{\mathrm{x}}^2\right)$ can be written as $44 {\mathrm{x}}^2\left({\mathrm{x}}^2-5\mathrm{x}-24\right)$ . 

Further, $\left({\mathrm{x}}^2-5\mathrm{x}-24\right)$ can be split into two factors, by splitting its middle term and using the identity $(\mathrm{x}+\mathrm{a})(\mathrm{x}+\mathrm{b})={\mathrm{x}}^2+(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}$.

Comparing these two, we get $(\mathrm{a}+\mathrm{b})=-5 \mathrm{and} \mathrm{ab}=-24$ 

From the above condition, we can get, $\mathrm{a}=-8 \mathrm{and} \mathrm{b}=3$

Now,

$\begin{array}{lcl} & & 44\left({\mathrm{x}}^4-5{\mathrm{x}}^3-24{\mathrm{x}}^2\right)÷11\mathrm{x}\left(\mathrm{x}-8\right) \\ & =& \frac{44\left({\mathrm{x}}^4-5{\mathrm{x}}^3-24{\mathrm{x}}^2\right)}{11\mathrm{x}\left(\mathrm{x}-8\right)} \\ & =& \frac{44 {\mathrm{x}}^2\left({\mathrm{x}}^2-5\mathrm{x}-24\right)}{11\mathrm{x}(\mathrm{x}-8)} \\ & =& \frac{44 {\mathrm{x}}^2({\mathrm{x}}^2-8\mathrm{x}+3\mathrm{x}-24)}{11\mathrm{x}(\mathrm{x}-8)} \\ & =& \frac{44 {\mathrm{x}}^2 [\mathrm{x}\left(\mathrm{x}-8\right)+3(\mathrm{x}-8\left)\right]}{11\mathrm{x}(\mathrm{x}-8)}\\ & =& \frac{{\displaystyle \stackrel{4}{\cancel{44}}}\times \mathrm{x}\times \cancel{\mathrm{x}}\cancel{(\mathrm{x}-8)}(\mathrm{x}+3)}{\cancel{11}\times \cancel{\mathrm{x}}\times \cancel{(\mathrm{x}-8)}} \\ & =& 4\mathrm{x}(\mathrm{x}+3)\end{array}$