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

Solve it by horner′s method and get the quotient. 2x^3 y+3xy−5x^2 y^2 +12÷(2x−4)=?

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

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

f(x)=2x^3 y−3xy−5x^2 y^2 +12 f(x)=(2x−4)(ax^2 +bx+c)+d =2ax^3 +(2b−4a)x^2 +(2c−4b)x+(d−4c) { ((a=^x^3 y)),((2b−4a=^x^2 −5y^2 ⇒b=((4y−5y^2 )/2))),((2c−4b=−3y⇒c=4y−5y^2 −(3/2)y)),((d−4c=12 ⇒d=10y−20y^2 +12)) :} f(x)=(2x−4)(yx^2 +(((4y−5y^2 )/2))x+((5/2)y−5y^2 ))+(12+10y−20y^2 )

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

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

P(x)=2x^3 y+3xy−5x^2 y^2 +12 D(x)=2x−4 •((P(x))/(D(x))) & ((P(x)/2)/(D(x)/2)) have same quotient but the latter has half of the remainder P(x)/2=x^3 y−(5/2)x^2 y^2 +(3/2)xy+6 D(x)/2=x−2 Now by synthetic division: 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))) Q(x)=(y)x^2 +(-(5/2)y^2 +2y)x+(-5y^2 +((11)/2)y) R=(-10y^2 +11y+6)×2=-20y^2 +22y+12

$${P}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} {y}+\mathrm{3}{xy}−\mathrm{5}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{12} \\ $$$${D}\left({x}\right)=\mathrm{2}{x}−\mathrm{4} \\ $$$$\bullet\frac{{P}\left({x}\right)}{{D}\left({x}\right)}\:\&\:\frac{{P}\left({x}\right)/\mathrm{2}}{{D}\left({x}\right)/\mathrm{2}}\:{have}\:{same}\:{quotient} \\ $$$$\:\:\:{but}\:{the}\:{latter}\:{has}\:{half}\:{of}\:{the}\: \\ $$$$\:\:\:{remainder} \\ $$$${P}\left({x}\right)/\mathrm{2}={x}^{\mathrm{3}} {y}−\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{xy}+\mathrm{6} \\ $$$${D}\left({x}\right)/\mathrm{2}={x}−\mathrm{2} \\ $$$${Now}\:{by}\:{synthetic}\:{division}: \\ $$$$\begin{array}{|c|c|c|}{\left.\mathrm{2}\right)}&\hline{{y}}&\hline{-\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} }&\hline{\frac{\mathrm{3}}{\mathrm{2}}{y}}&\hline{\mathrm{6}}\\{}&\hline{}&\hline{\mathrm{2}{y}}&\hline{-\mathrm{5}{y}^{\mathrm{2}} +\mathrm{4}{y}}&\hline{-\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}}\\{}&\hline{{y}}&\hline{-\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{2}{y}}&\hline{-\mathrm{5}{y}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{2}}{y}}&\hline{-\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}+\mathrm{6}}\\\hline\end{array} \\ $$$${Q}\left({x}\right)=\left({y}\right){x}^{\mathrm{2}} +\left(-\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{2}{y}\right){x}+\left(-\mathrm{5}{y}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{2}}{y}\right) \\ $$$${R}=\left(-\mathrm{10}{y}^{\mathrm{2}} +\mathrm{11}{y}+\mathrm{6}\right)×\mathrm{2}=-\mathrm{20}{y}^{\mathrm{2}} +\mathrm{22}{y}+\mathrm{12} \\ $$

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