$\displaystyle{27}{{x}}^{{3}}{{y}}^{{2}}+{12}{{x}}^{{2}}{{y}}^{{5}}={3}{x}{{y}}^{{2}}{\left({9}{{x}}^{{2}}+{4}{x}{{y}}^{{3}}\right)}\usepackage{cmbright}\Huge\$ and $\displaystyle{27}{{x}}^{{3}}{{y}}^{{2}}+{12}{{x}}^{{2}}{{y}}^{{5}}={3}{{x}}^{{2}}{{y}}^{{2}}{\left({9}{x}+{4}{{y}}^{{3}}\right)}\usepackage{cmbright}\Huge\$ , so both factorizations are correct.