Question Number 178413 by mathlove last updated on 16/Oct/22
$$\left(\mathrm{2}{x}^{\mathrm{3}} {y}+\mathrm{3}{xy}−\mathrm{5}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{12}\right)\boldsymbol{\div}\left(\mathrm{2}{x}−\mathrm{4}\right) \\ $$$$ \\ $$Divide it using compound division
Commented by Rasheed.Sindhi last updated on 17/Oct/22
$${What}'{s}\:{meant}\:{by}\:“{compound}\:{division}''? \\ $$
Commented by mathlove last updated on 18/Oct/22
$${Harner}\:{divaided} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Oct/22
![(2x^3 y+3xy−5x^2 y^2 +12)÷(2x−4) =(x^3 y−(5/2)x^2 y^2 +(3/2)xy+6)÷(x−2) [In order to obtain the divisor of the form x−α , we′ve divided both dividend & divisor by 2 ] determinant (((2)),y,(−(5/2)y^2 ),((3/2)y),6),( , ,(2y),(−5y^2 +4y),(−10y^2 +11y)),( ,y,(−(5/2)y^2 +2y),(−5y^2 +((11)/2)y),(−10y^2 +11y+6))) x^2 y+x(−(5/2)y^2 +2y)−5y^2 +((11)/2)y+((−10y^2 +11y+6)/(x−2)) ◂Ans](https://www.tinkutara.com/question/Q178549.png)
$$\left(\mathrm{2}{x}^{\mathrm{3}} {y}+\mathrm{3}{xy}−\mathrm{5}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{12}\right)\boldsymbol{\div}\left(\mathrm{2}{x}−\mathrm{4}\right) \\ $$$$=\left({x}^{\mathrm{3}} {y}−\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{xy}+\mathrm{6}\right)\boldsymbol{\div}\left({x}−\mathrm{2}\right) \\ $$$$\left[{In}\:{order}\:{to}\:{obtain}\:{the}\:{divisor}\:{of}\:{the}\right. \\ $$$$\:\:{form}\:{x}−\alpha\:,\:{we}'{ve}\:{divided}\:{both} \\ $$$$\left.\:\:\:{dividend}\:\&\:{divisor}\:{by}\:\mathrm{2}\:\right] \\ $$$$\begin{array}{|c|c|c|}{\left.\mathrm{2}\right)}&\hline{{y}}&\hline{−\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} }&\hline{\frac{\mathrm{3}}{\mathrm{2}}{y}}&\hline{\mathrm{6}}\\{\:}&\hline{\:}&\hline{\mathrm{2}{y}}&\hline{−\mathrm{5}{y}^{\mathrm{2}} +\mathrm{4}{y}}&\hline{−\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}}\\{\:}&\hline{{y}}&\hline{−\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{2}{y}}&\hline{−\mathrm{5}{y}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{2}}{y}}&\hline{−\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}+\mathrm{6}}\\\hline\end{array}\: \\ $$$${x}^{\mathrm{2}} {y}+{x}\left(−\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{2}{y}\right)−\mathrm{5}{y}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{2}}{y}+\frac{−\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}+\mathrm{6}}{{x}−\mathrm{2}}\:\:\blacktriangleleft\boldsymbol{\mathrm{Ans}} \\ $$
Commented by mathlove last updated on 18/Oct/22
$${thanks}\:{a}\:{lot}\:{my}\:{teacher} \\ $$