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Question Number 175483 by sciencestudent last updated on 31/Aug/22

$$Solveitby{horner}^{\prime }smethodandget$$$$thequotient.$$$$2x^{3}y+3xy-5x^2{y}^2+12\mathbf{÷}(2x-4)=?$$

Answered by Ar Brandon last updated on 31/Aug/22

$$f\left(x\right)=2x^3\mathrm{y}-3x\mathrm{y}-5x^2{\mathrm{y}}^2+12$$$$f\left(x\right)=(2x-4)({ax}^2+bx+c)+d$$$$=2{ax}^3+(2b-4a)x^2+(2c-4b)x+(d-4c)$$$$\{\begin{array}{l}a\stackrel{x^3}{=}\mathrm{y}\\ 2b-4a\stackrel{x^2}{=}-{5\mathrm{y}}^2\Rightarrow b=\frac{4\mathrm{y}-{5\mathrm{y}}^2}{2}\\ 2c-4b=-3\mathrm{y}\Rightarrow c=4\mathrm{y}-{5\mathrm{y}}^2-\frac{3}{2}\mathrm{y}\\ d-4c=12\Rightarrow d=10\mathrm{y}-{20\mathrm{y}}^2+12\end{array}$$$$f\left(x\right)=(2x-4)(\mathrm{y}{x}^2+\left(\frac{4\mathrm{y}-{5\mathrm{y}}^2}{2}\right)x+(\frac{5}{2}\mathrm{y}-{5\mathrm{y}}^2))+(12+10\mathrm{y}-{20\mathrm{y}}^2)$$

Answered by Rasheed.Sindhi last updated on 31/Aug/22

$$P\left(x\right)=2x^{3}y+3xy-5x^2{y}^2+12$$$$D\left(x\right)=2x-4$$$$\bullet \frac{P\left(x\right)}{D\left(x\right)}\mathrm{\&}\frac{P\left(x\right)/2}{D\left(x\right)/2}havesamequotient$$$$butthelatterhashalfofthe$$$$remainder$$$$P\left(x\right)/2=x^{3}y-\frac{5}{2}{x}^2{y}^2+\frac{3}{2}xy+6$$$$D\left(x\right)/2=x-2$$$$Nowbysyntheticdivision:$$$$\overline{)\begin{array}{ccccc}2)& y& -\frac{5}{2}{y}^2& \frac{3}{2}y& 6\\ & & 2y& -5y^2+4y& -10y^2+11y\\ & y& -\frac{5}{2}{y}^2+2y& -5y^2+\frac{11}{2}y& -10y^2+11y+6\end{array}}$$$$Q\left(x\right)=\left(y\right)x^2+(-\frac{5}{2}{y}^2+2y)x+(-5y^2+\frac{11}{2}y)$$$$R=(-10y^2+11y+6)\times 2=-20y^2+22y+12$$

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