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Question Number 26721 by shiv15031973@gmail.com last updated on 02/Jan/18

−2y(y−12)(y−1)or(y−12)(−2y^2 −2y)  both are same.

$$−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right){or}\left({y}−\mathrm{12}\right)\left(−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}\right)\:\:{both}\:{are}\:{same}. \\ $$

Commented by Rasheed.Sindhi last updated on 28/Dec/17

Do you mean ((−2y(y−12)(y−1))/((y−12)(−2y^2 −2y))) ?  If you mean so  ((−2y(y−12)(y−1))/((y−12)(−2y^2 −2y)))  =((−2y(y−12)(y−1))/(−2y(y−12)(y+1)))  =((y−1)/(y+1))

$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:\frac{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right)}{\left({y}−\mathrm{12}\right)\left(−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}\right)}\:? \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{so} \\ $$$$\frac{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right)}{\left({y}−\mathrm{12}\right)\left(−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}\right)} \\ $$$$=\frac{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}−\mathrm{1}\right)}{−\mathrm{2}{y}\left({y}−\mathrm{12}\right)\left({y}+\mathrm{1}\right)} \\ $$$$=\frac{{y}−\mathrm{1}}{{y}+\mathrm{1}} \\ $$

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