Correct. This expression is a perfect square trinomial because $\displaystyle{36}{{x}}^{{2}}={{\left({6}{x}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ , $\displaystyle{1}={{1}}^{{2}}\usepackage{cmbright}\Huge\$ , and $\displaystyle{12}{x}={2}\cdot{6}{x}\cdot{1}\usepackage{cmbright}\Huge\$ . Therefore, $\displaystyle{36}{{x}}^{{2}}+{12}{x}+{1}={{\left({6}{x}+{1}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ . The missing value is 6.