Incorrect. If $\displaystyle{{x}}^{{2}}+{14}{x}+{49}\usepackage{cmbright}\Huge\$ is a perfect square trinomial, then $\displaystyle{{x}}^{{2}}+{14}{x}+{49}={\left({x}+{r}\right)}{\left({x}+{r}\right)}\usepackage{cmbright}\Huge\$ . What must r be?