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\\n
Commented byRasheed.Sindhi last updated on 17/Oct/22
$${What}'{s}\:{meant}\:{by}\:\natural{compound}\:{division}\varepsilon? \\ $$
Commented bymathlove 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](Q178549.png)
$$\left(\mathrm{2}{x}^{\mathrm{3}} {y}+\mathrm{3}{xy}−\mathrm{5}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{12}\right)\boldsymbol{\div}\left(\mathrm{2}{x}−\mathrm{4}\right) \\ $$ $$=\left({x}^{\mathrm{3}} {y}−\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{xy}+\mathrm{6}\right)\boldsymbol{\div}\left({x}−\mathrm{2}\right) \\ $$ $$\color{mathbrown}{\left[}{\color{mathbrown}{I}\color{mathbrown}{n}}\color{mathbrown}{\:}{\color{mathbrown}{o}\color{mathbrown}{r}\color{mathbrown}{d}\color{mathbrown}{e}\color{mathbrown}{r}}\color{mathbrown}{\:}{\color{mathbrown}{t}\color{mathbrown}{o}}\color{mathbrown}{\:}{\color{mathbrown}{o}\color{mathbrown}{b}\color{mathbrown}{t}\color{mathbrown}{a}\color{mathbrown}{i}\color{mathbrown}{n}}\color{mathbrown}{\:}{\color{mathbrown}{t}\color{mathbrown}{h}\color{mathbrown}{e}}\color{mathbrown}{\:}{\color{mathbrown}{d}\color{mathbrown}{i}\color{mathbrown}{v}\color{mathbrown}{i}\color{mathbrown}{s}\color{mathbrown}{o}\color{mathbrown}{r}}\color{mathbrown}{\:}{\color{mathbrown}{o}\color{mathbrown}{f}}\color{mathbrown}{\:}{\color{mathbrown}{t}\color{mathbrown}{h}\color{mathbrown}{e}}\right. \\ $$ $$\color{mathbrown}{\:}\color{mathbrown}{\:}{\color{mathbrown}{f}\color{mathbrown}{o}\color{mathbrown}{r}\color{mathbrown}{m}}\color{mathbrown}{\:}{\color{mathbrown}{x}}\color{mathbrown}{−}\color{mathbrown}{\alpha}\color{mathbrown}{\:},\color{mathbrown}{\:}{\color{mathbrown}{w}\color{mathbrown}{e}}\color{mathbrown}{'}{\color{mathbrown}{v}\color{mathbrown}{e}}\color{mathbrown}{\:}{\color{mathbrown}{d}\color{mathbrown}{i}\color{mathbrown}{v}\color{mathbrown}{i}\color{mathbrown}{d}\color{mathbrown}{e}\color{mathbrown}{d}}\color{mathbrown}{\:}{\color{mathbrown}{b}\color{mathbrown}{o}\color{mathbrown}{t}\color{mathbrown}{h}} \\ $$ $$\left.\color{mathbrown}{\:}\color{mathbrown}{\:}\color{mathbrown}{\:}{\color{mathblue}{d}\color{mathblue}{i}\color{mathblue}{v}\color{mathblue}{i}\color{mathblue}{d}\color{mathblue}{e}\color{mathblue}{n}\color{mathblue}{d}}\color{mathblue}{\:}\&\color{mathblue}{\:}{\color{mathblue}{d}\color{mathblue}{i}\color{mathblue}{v}\color{mathblue}{i}\color{mathblue}{s}\color{mathblue}{o}\color{mathblue}{r}}\:{\color{mathbrown}{b}\color{mathbrown}{y}}\color{mathbrown}{\:}\mathrm{\color{mathbrown}{2}}\color{mathbrown}{\:}\color{mathbrown}{\right]} \\ $$ $$\begin{array}{|c|c|c|}{\left.\mathrm{2}\right)}&\hline{{y}}&\hline{−\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} }&\hline{\frac{\mathrm{3}}{\mathrm{2}}{y}}&\hline{\mathrm{6}}\\{\:}&\hline{\:}&\hline{\mathrm{2}{y}}&\hline{−\mathrm{5}{y}^{\mathrm{2}} +\mathrm{4}{y}}&\hline{−\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}}\\{\:}&\hline{{y}}&\hline{−\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{2}{y}}&\hline{−\mathrm{5}{y}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{2}}{y}}&\hline{−\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}+\mathrm{6}}\\\hline\end{array}\: \\ $$ $${\color{mathred}{x}}^{\mathrm{\color{mathred}{2}}} {y}+{\color{mathred}{x}}\left(−\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{2}{y}\right)−\mathrm{5}{y}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{2}}{y}+\frac{−\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}+\mathrm{6}}{{x}−\mathrm{2}}\:\:\color{mathred}{\blacktriangleleft}\boldsymbol{\mathrm{\color{mathred}{A}\color{mathred}{n}\color{mathred}{s}}} \\ $$
Commented bymathlove last updated on 18/Oct/22
$${thanks}\:{a}\:{lot}\:{my}\:{teacher} \\ $$