Factorize-60x-3-y-115x-2-y-2-120xy-3-

Question Number 41710 by Tawa1 last updated on 11/Aug/18

$$\mathrm{Factorize}:\:\:\mathrm{60x}^{\mathrm{3}} \mathrm{y}\:+\:\mathrm{115x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:−\:\mathrm{120xy}^{\mathrm{3}} \\ $$

Answered by alex041103 last updated on 11/Aug/18

60x^3 y + 115x^2 y^2 − 120xy^3 = =5xy(12x^2 +23xy−24y^2 ) Then let′sfind x_(1,2) for (12)x^2 +(23y)x+(−24y^2 )=0 ⇒x_(1,2) =((−23y±(√(529y^2 +1152y^2 )))/(24))= =((±(√(1681))−23)/(24))y=((−23±41)/(24))y=−(8/3)y or (3/4)y ⇒(12x^2 +23xy−24y^2 )=12(x+(8/3)y)(x−(3/4)y) =(3x+8y)(4x−3y) ⇒60x^3 y + 115x^2 y^2 − 120xy^3 = =5xy(3x+8y)(4x−3y)

$$\mathrm{60x}^{\mathrm{3}} \mathrm{y}\:+\:\mathrm{115x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:−\:\mathrm{120x}{y}^{\mathrm{3}} = \\ $$$$=\mathrm{5}{xy}\left(\mathrm{12}{x}^{\mathrm{2}} +\mathrm{23}{xy}−\mathrm{24}{y}^{\mathrm{2}} \right) \\ $$$${Then}\:{let}'{sfind}\:{x}_{\mathrm{1},\mathrm{2}} {for} \\ $$$$\left(\mathrm{12}\right){x}^{\mathrm{2}} +\left(\mathrm{23}{y}\right){x}+\left(−\mathrm{24}{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{23}{y}\pm\sqrt{\mathrm{529}{y}^{\mathrm{2}} +\mathrm{1152}{y}^{\mathrm{2}} }}{\mathrm{24}}= \\ $$$$=\frac{\pm\sqrt{\mathrm{1681}}−\mathrm{23}}{\mathrm{24}}{y}=\frac{−\mathrm{23}\pm\mathrm{41}}{\mathrm{24}}{y}=−\frac{\mathrm{8}}{\mathrm{3}}{y}\:{or}\:\frac{\mathrm{3}}{\mathrm{4}}{y} \\ $$$$\Rightarrow\left(\mathrm{12}{x}^{\mathrm{2}} +\mathrm{23}{xy}−\mathrm{24}{y}^{\mathrm{2}} \right)=\mathrm{12}\left({x}+\frac{\mathrm{8}}{\mathrm{3}}{y}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{4}}{y}\right) \\ $$$$=\left(\mathrm{3}{x}+\mathrm{8}{y}\right)\left(\mathrm{4}{x}−\mathrm{3}{y}\right) \\ $$$$\Rightarrow\mathrm{60x}^{\mathrm{3}} \mathrm{y}\:+\:\mathrm{115x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:−\:\mathrm{120x}{y}^{\mathrm{3}} = \\ $$$$=\mathrm{5}{xy}\left(\mathrm{3}{x}+\mathrm{8}{y}\right)\left(\mathrm{4}{x}−\mathrm{3}{y}\right) \\ $$

Commented by Tawa1 last updated on 11/Aug/18

I really appreciate sir. God bless you sir

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by ajfour last updated on 11/Aug/18

=5xy(12x^2 +23xy−24y^2 ) =5xy(12x^2 +32xy−9xy−24y^2 ) =5xy[4x(3x+8y)−3y(3x+8y)] =5xy(3x+8y)(4x−3y) .

$$=\mathrm{5}{xy}\left(\mathrm{12}{x}^{\mathrm{2}} +\mathrm{23}{xy}−\mathrm{24}{y}^{\mathrm{2}} \right) \\ $$$$=\mathrm{5}{xy}\left(\mathrm{12}{x}^{\mathrm{2}} +\mathrm{32}{xy}−\mathrm{9}{xy}−\mathrm{24}{y}^{\mathrm{2}} \right) \\ $$$$=\mathrm{5}{xy}\left[\mathrm{4}{x}\left(\mathrm{3}{x}+\mathrm{8}{y}\right)−\mathrm{3}{y}\left(\mathrm{3}{x}+\mathrm{8}{y}\right)\right] \\ $$$$=\mathrm{5}{xy}\left(\mathrm{3}{x}+\mathrm{8}{y}\right)\left(\mathrm{4}{x}−\mathrm{3}{y}\right)\:. \\ $$

Commented by Tawa1 last updated on 11/Aug/18

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Facebook Tweet Pin

Factorize-60x-3-y-115x-2-y-2-120xy-3-

Leave a Reply Cancel reply