Incorrect. Notice that the coefficient of $\displaystyle{{x}}^{{2}}{\left({1}={{1}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ and the constant term $\displaystyle{\left({9}={{3}}^{{2}}={{\left(-{3}\right)}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ are both squares. This particular polynomial can be factored as the square of a binomial $\displaystyle{{\left({x}+{r}\right)}}^{{2}}={\left({x}+{r}\right)}{\left({x}+{r}\right)}\usepackage{cmbright}\Huge\$ or $\displaystyle{{\left({x}-{r}\right)}}^{{2}}={\left({x}-{r}\right)}{\left({x}-{r}\right)}\usepackage{cmbright}\Huge\$ . What must r be?