$\left(5{\mathrm{x}}^2-25\mathrm{x}+20\right)÷(\mathrm{x}-1)=\mathrm{\_\_\_\_\_\_\_\_}$

$\left(5{\mathrm{x}}^2-25\mathrm{x}+20\right)$ can be written as $5\left({\mathrm{x}}^2-5\mathrm{x}+4\right)$. 

Further, $\left({\mathrm{x}}^2-5\mathrm{x}+4\right)$ 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 $(\mathrm{a}+\mathrm{b})=-5 \mathrm{and} \mathrm{ab}=4$ 

From the above condition, we can get, $\mathrm{a}=-4 \mathrm{and} \mathrm{b}=-1$

$\begin{array}{rcl}\left(5{\mathrm{x}}^2-25\mathrm{x}+20\right)÷(\mathrm{x}-1)& =& \frac{\left(5{\mathrm{x}}^2-25\mathrm{x}+20\right)}{(\mathrm{x}-1)}\\ & =& \frac{5\left({\mathrm{x}}^2-5\mathrm{x}+4\right)}{(\mathrm{x}-1)}\\ & =& \frac{5\left({\mathrm{x}}^2-4\mathrm{x}-\mathrm{x}+4\right)}{\mathrm{x}-1}\\ & =& \frac{5(\mathrm{x}-4)(\mathrm{x}-1)}{\mathrm{x}-1}\\ & =& \frac{5(\mathrm{x}-4)\cancel{(\mathrm{x}-1)}}{\cancel{\mathrm{x}-1}}\\ & =& 5(\mathrm{x}-4)\end{array}$