$\displaystyle{\left({2}{x}+{1}\right)}{\left({2}{x}+{2}\right)}={4}{{x}}^{{2}}+{6}{x}+{2}\usepackage{cmbright}\Huge\$ , so this answer is incorrect. Notice that $\displaystyle{2}{{x}}^{{2}}+{6}{x}+{4}={2}{\left({{x}}^{{2}}+{3}{x}+{2}\right)}\usepackage{cmbright}\Huge\$ . Can you think of a factorization of $\displaystyle{{x}}^{{2}}+{3}{x}+{2}\usepackage{cmbright}\Huge\$ ? Are there integers r and s so that $\displaystyle{r}{s}={2}\usepackage{cmbright}\Huge\$ and $\displaystyle{r}+{s}={3}\usepackage{cmbright}\Huge\$ ?