Correct. Factoring the numerator and denominator and pulling out expressions equal to 1, you end up with $\displaystyle\frac{{{\left({c}+{2}\right)}{\left({c}+{2}\right)}}}{{{\left({2}{c}-{5}\right)}{\left({c}+{2}\right)}}}\cdot\frac{{{c}+{5}}}{{{c}{\left({c}+{2}\right)}}}=\frac{{{\left({c}+{2}\right)}{\left({c}+{2}\right)}}}{{{\left({c}+{2}\right)}{\left({c}+{2}\right)}}}\cdot\frac{{{c}+{5}}}{{{c}{\left({2}{c}-{5}\right)}}}=\frac{{{c}+{5}}}{{{c}{\left({2}{c}-{5}\right)}}}\usepackage{cmbright}\Huge\$ . Since the denominator of the expression will be 0 when $\displaystyle{c}=-{2}\usepackage{cmbright}\Huge\$ , 0, and 2.5, those values need to be excluded.