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Question Number 175346 by sciencestudent last updated on 27/Aug/22

Solve it by horner′s method.  (2x^3 y^2 +3x^2 y−4x+5y−12)÷(x−3)

$${Solve}\:{it}\:{by}\:{horner}'{s}\:{method}. \\ $$$$\left(\mathrm{2}{x}^{\mathrm{3}} {y}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} {y}−\mathrm{4}{x}+\mathrm{5}{y}−\mathrm{12}\right)\boldsymbol{\div}\left({x}−\mathrm{3}\right) \\ $$

Answered by cortano1 last updated on 28/Aug/22

   determinant ((•,(2y^2 ),(3y),(−4),(5y−12)),(3,∗,(6y^2 ),(18y^2 +9y),(54y^2 +27y−12)),(,(2y^2 ),(6y^2 +3y),(18y^2 +9y−4),(54y^2 +32y−24_(remainder) )))

$$\:\:\begin{array}{|c|c|c|}{\bullet}&\hline{\mathrm{2}{y}^{\mathrm{2}} }&\hline{\mathrm{3}{y}}&\hline{−\mathrm{4}}&\hline{\mathrm{5}{y}−\mathrm{12}}\\{\mathrm{3}}&\hline{\ast}&\hline{\mathrm{6}{y}^{\mathrm{2}} }&\hline{\mathrm{18}{y}^{\mathrm{2}} +\mathrm{9}{y}}&\hline{\mathrm{54}{y}^{\mathrm{2}} +\mathrm{27}{y}−\mathrm{12}}\\{}&\hline{\mathrm{2}{y}^{\mathrm{2}} }&\hline{\mathrm{6}{y}^{\mathrm{2}} +\mathrm{3}{y}}&\hline{\mathrm{18}{y}^{\mathrm{2}} +\mathrm{9}{y}−\mathrm{4}}&\hline{\underset{{remainder}} {\underbrace{\mathrm{54}{y}^{\mathrm{2}} +\mathrm{32}{y}−\mathrm{24}}}}\\\hline\end{array} \\ $$

Commented by sciencestudent last updated on 30/Aug/22

what is the quotient?

$${what}\:{is}\:{the}\:{quotient}? \\ $$

Commented by sciencestudent last updated on 30/Aug/22

quotient?

$${quotient}? \\ $$

Commented by sciencestudent last updated on 30/Aug/22

Please help me  mr W!

$${Please}\:{help}\:{me}\:\:{mr}\:{W}! \\ $$

Commented by cortano1 last updated on 30/Aug/22

the quotient is 2y^2 x^2 +(6y^2 +3y)x+18y^2 +9y−4

$${the}\:{quotient}\:{is}\:\mathrm{2}{y}^{\mathrm{2}} {x}^{\mathrm{2}} +\left(\mathrm{6}{y}^{\mathrm{2}} +\mathrm{3}{y}\right){x}+\mathrm{18}{y}^{\mathrm{2}} +\mathrm{9}{y}−\mathrm{4} \\ $$

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