Both x and 5 should be multiplied by $\displaystyle{2}{x}\usepackage{cmbright}\Huge\$ . Recall that $\displaystyle{x}\cdot{2}{x}={2}{{x}}^{{2}}\usepackage{cmbright}\Huge\$ , but $\displaystyle{5}\cdot{2}{x}\ne{10}{{x}}^{{2}}\usepackage{cmbright}\Huge\$ .