Question Number 26721 by shiv15031973@gmail.com last updated on 02/Jan/18
$$−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right){or}\left({y}−\mathrm{12}\right)\left(−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}\right)\:\:{both}\:{are}\:{same}. \\ $$
Commented by Rasheed.Sindhi last updated on 28/Dec/17
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:\frac{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right)}{\left({y}−\mathrm{12}\right)\left(−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}\right)}\:? \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{so} \\ $$$$\frac{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right)}{\left({y}−\mathrm{12}\right)\left(−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}\right)} \\ $$$$=\frac{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right)}{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}+\mathrm{1}\right)} \\ $$$$=\frac{{y}−\mathrm{1}}{{y}+\mathrm{1}} \\ $$