Incorrect. Notice that the coefficient of $\displaystyle{{a}}^{{2}}{\left({1}={{1}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ and the constant term $\displaystyle{\left({16}={{4}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ are both squares. This particular polynomial can be factored as the square of a binomial $\displaystyle{{\left({a}+{r}\right)}}^{{2}}={\left({a}+{r}\right)}{\left({a}+{r}\right)}\usepackage{cmbright}\Huge\$ . What must r be?