Correct. This expression can be rewritten as $\displaystyle\frac{{{3}{\left({{c}}^{{2}}+{1}\right)}}}{{{9}{c}}}\cdot\frac{{{c}{\left({{c}}^{{2}}+{5}\right)}}}{{{{c}}^{{2}}+{1}}}\usepackage{cmbright}\Huge\$ . Since there are no values for c that will make $\displaystyle{{c}}^{{2}}+{1}\usepackage{cmbright}\Huge\$ and $\displaystyle{{c}}^{{2}}+{5}\usepackage{cmbright}\Huge\$ equal 0, $\displaystyle{c}={0}\usepackage{cmbright}\Huge\$ is the only value that needs to be excluded. This expression simplifies to $\displaystyle\frac{{{{c}}^{{2}}+{5}}}{{3}}\usepackage{cmbright}\Huge\$ .