Incorrect. Notice that $\displaystyle{{b}}^{{2}}+{2}{b}+{1}\usepackage{cmbright}\Huge\$ and $\displaystyle{{b}}^{{2}}-{1}\usepackage{cmbright}\Huge\$ are both special products of binomials; does that help you factor this expression? Remember that the domain must exclude any values for b that make the denominator have a value of 0.