You probably recognized that $\displaystyle{{2}}^{{4}}={16}\usepackage{cmbright}\Huge\$ and $\displaystyle{{3}}^{{4}}={81}\usepackage{cmbright}\Huge\$ . However, $\displaystyle{{\left({2}{x}+{3}\right)}}^{{4}}\usepackage{cmbright}\Huge\$ is not equal to the given trinomial. A trinomial in the form $\displaystyle{{r}}^{{2}}+{2}{r}{s}+{{s}}^{{2}}\usepackage{cmbright}\Huge\$ can be factored as $\displaystyle{{\left({r}+{s}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ . Remember that $\displaystyle{16}{{x}}^{{2}}={{\left({4}{x}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ and $\displaystyle{81}={{9}}^{{2}}\usepackage{cmbright}\Huge\$ .