$\displaystyle{\left({x}+{1}\right)}{\left({x}+{2}\right)}={{x}}^{{2}}+{3}{x}+{2}\usepackage{cmbright}\Huge\$ , so this answer is incorrect. If the polynomial can be factored, the factorization will "look like" $\displaystyle{2}{{x}}^{{2}}+{3}{x}+{1}={\left({2}{x}+{r}\right)}{\left({x}+{s}\right)}\usepackage{cmbright}\Huge\$ (since the coefficient of $\displaystyle{{x}}^{{2}}\usepackage{cmbright}\Huge\$ is 2). Are there integers r and s so that $\displaystyle{r}{s}={1}\usepackage{cmbright}\Huge\$ and $\displaystyle{r}+{2}{s}={3}\usepackage{cmbright}\Huge\$ ?