Answer
$\dfrac{n^2+4n-5}{n+5} =n-1; n \ne -5$
Work Step by Step
Given: $\dfrac{n^2+4n-5}{n+5}$
Consider denominator $n+5=0$ to check the exclude value which makes the denominator $0$.
$n+5=0$
Thus, $n=-5$
$\dfrac{n^2+4n-5}{n+5} = \dfrac{(n+5)(n-1)}{(n+5)}$
Hence, $\dfrac{n^2+4n-5}{n+5} =n-1; n \ne -5$
