You may have incorrectly simplified the term $\displaystyle{2}{{x}}^{{0}}\usepackage{cmbright}\Huge\$ . The expression $\displaystyle{{x}}^{{0}}\usepackage{cmbright}\Huge\$ has a value of 1 if $\displaystyle{x}\ne{0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{2}{{x}}^{{0}}={2}{\left({1}\right)}={2}\usepackage{cmbright}\Huge\$ . The correct answer is $\displaystyle{{x}}^{{3}}+{5}{{x}}^{{2}}+{2}\usepackage{cmbright}\Huge\$ .