Answer
$\dfrac{1}{k+4}$; $k\ne2$ and $k\ne-4$
Work Step by Step
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 $k\ne-4$
