Incorrect. Notice that the coefficient of $\displaystyle{{x}}^{{2}}{\left({1}={{1}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ and the constant term $\displaystyle{\left({81}={{9}}^{{2}}={{\left(-{9}\right)}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ are both squares. Could this polynomial be a perfect square trinomial? If so, it equals $\displaystyle{{\left({x}+{r}\right)}}^{{2}}={\left({x}+{r}\right)}{\left({x}+{r}\right)}\usepackage{cmbright}\Huge\$ . What must r be?