## Algebra 1

$\dfrac{1}{k+4}$; $k\ne2$ and $k\ne-4$
Given: $\dfrac{k-2}{k^2+2k-8}$ Need to find the common factors of the given expression. $\dfrac{k-2}{k^2+2k-8}=\dfrac{k-2}{(k-2)(k+4)}$ $=\dfrac{1}{k+4}$ If we put $k=2$ and $k=-4$ then denominator becomes zero, which cannot be possible. After simplification, we get $=\dfrac{1}{k+4}$; $k\ne2$ and $k\ne-4$