Correct. Noticing $\displaystyle{\left({x}-{3}\right)}\usepackage{cmbright}\Huge\$ is a factor of $\displaystyle{{x}}^{{2}}-{3}{x}\usepackage{cmbright}\Huge\$ and $\displaystyle{2}{x}+{12}\usepackage{cmbright}\Huge\$ , you found that the expression equals $\displaystyle{x}{\left({x}-{3}\right)}-{2}{\left({x}-{3}\right)}\usepackage{cmbright}\Huge\$ , which equals $\displaystyle{\left({x}-{2}\right)}{\left({x}-{3}\right)}\usepackage{cmbright}\Huge\$ .