Correct. $\displaystyle{{x}}^{{2}}+{2}{x}+{3}{x}+{6}={x}{\left({x}+{2}\right)}+{3}{\left({x}+{2}\right)}\usepackage{cmbright}\Huge\$ , so

$\displaystyle{\left({x}+{2}\right)}{\left({x}+{3}\right)}={0}\usepackage{cmbright}\Huge\$

$\displaystyle{x}+{2}={0}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}+{3}={0}\usepackage{cmbright}\Huge\$

$\displaystyle{x}=-{2}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}=-{3}\usepackage{cmbright}\Huge\$