Incorrect. Notice that the coefficient of $\displaystyle{{x}}^{{2}}{\left({1}={{1}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ and the constant term $\displaystyle{\left({64}={{8}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ are both squares. Could this polynomial be a perfect square trinomial? If so, it equals $\displaystyle{{\left({a}+{r}\right)}}^{{2}}={\left({a}+{r}\right)}{\left({a}+{r}\right)}\usepackage{cmbright}\Huge\$ or $\displaystyle{{\left({a}-{r}\right)}}^{{2}}={\left({a}-{r}\right)}{\left({a}-{r}\right)}\usepackage{cmbright}\Huge\$ . What must a and r equal?