Find the value of $({\mathrm{m}}^2-15\mathrm{m}-16)÷(\mathrm{m}-16)$ at $\mathrm{m}=2$.

 $({\mathrm{m}}^2-15\mathrm{m}-16)$ 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 $(\mathrm{a}+\mathrm{b})=-15 \mathrm{and} \mathrm{ab}=-16$ 

From the above condition, we can get, $\mathrm{a}=-16 \mathrm{and} \mathrm{b}=1$

$\begin{array}{rcl}\frac{\left({\mathrm{m}}^2-15\mathrm{m}-16\right)}{\left(\mathrm{m}-16\right)}& =& \frac{\left({\mathrm{m}}^2-15\mathrm{m}-16\right)}{\mathrm{m}-16}\\ & =& \frac{\left({\mathrm{m}}^2-16\mathrm{m}+\mathrm{m}-16\right)}{\mathrm{m}-16}\\ & =& \frac{\mathrm{m}(\mathrm{m}-16)+1(\mathrm{m}-16)}{\mathrm{m}-16}\\ & =& \frac{\cancel{(\mathrm{m}-16)}(\mathrm{m}+1)}{\cancel{\mathrm{m}-16}}\\ & =& \mathrm{m}+1\end{array}$

Now, for $\mathrm{m}=2$,

 $\mathrm{m}+1=\left(2\right)+1$

$=3$