Correct. You recognized that $\displaystyle\frac{{{3}{x}+{3}}}{{{3}{{x}}^{{2}}-{9}{x}-{12}}}=\frac{{{3}{\left({x}+{1}\right)}}}{{{3}{\left({x}+{1}\right)}{\left({x}-{4}\right)}}}\usepackage{cmbright}\Huge\$ . Rewriting with factors of 1, you have $\displaystyle\frac{{{3}{\left({x}+{1}\right)}}}{{{3}{\left({x}+{1}\right)}}}\cdot\frac{{1}}{{{x}-{4}}}={1}\cdot\frac{{1}}{{{\left({x}-{4}\right)}}}=\frac{{1}}{{{\left({x}-{4}\right)}}}\usepackage{cmbright}\Huge\$ .