Solve:

$\left({\mathrm{m}}^2-14\mathrm{m}-32\right)÷(\mathrm{m}-16)(\mathrm{m}+2)$

$\left({\mathrm{m}}^2-14\mathrm{m}-32\right)$ 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 $(\mathrm{a}+\mathrm{b})=-14 \mathrm{and} \mathrm{ab}=-32$ .

From the above condition, we can get, $\mathrm{a}=-16 \mathrm{and} \mathrm{b}=2$

$\begin{array}{rcl}\therefore \frac{\left({\mathrm{m}}^2-14\mathrm{m}-32\right)}{(\mathrm{m}-16)(\mathrm{m}+2)}& =& \frac{\left({\mathrm{m}}^2-14\mathrm{m}-32\right)}{(\mathrm{m}-16)(\mathrm{m}+2)}\\ & =& \frac{\left({\mathrm{m}}^2-16\mathrm{m}+2\mathrm{m}-32\right)}{(\mathrm{m}-16)(\mathrm{m}+2)}\\ & =& \frac{\mathrm{m}(\mathrm{m}-16)+2(\mathrm{m}-16)}{(\mathrm{m}-16)(\mathrm{m}+2)}\\ & =& \frac{(\mathrm{m}-16)(\mathrm{m}+2)}{(\mathrm{m}-16)(\mathrm{m}+2)}\\ & =& \frac{\cancel{(\mathrm{m}-16)}\cancel{(\mathrm{m}+2)}}{\cancel{(\mathrm{m}-16)}\cancel{(\mathrm{m}+2)}}\\ & =& 1\\ & & \end{array}$