Question Number 98859 by peter frank last updated on 16/Jun/20

Commented by mr W last updated on 16/Jun/20

$${p}\left({x}\right)={q}\left({x}\right)\left({x}−\mathrm{1}\right) \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{1}+{c}−\mathrm{2}{c}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{c}=\mathrm{5} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Jun/20

$$\begin{pmatrix}{\left.\mathrm{1}\right)}&{\mathrm{1}}&{\:\:\:{c}}&{−\mathrm{2}{c}}&{\:\:\:\:\:\mathrm{4}}\\{}&{}&{\:\:\:\mathrm{1}}&{\:\:{c}+\mathrm{1}}&{−{c}+\mathrm{1}}\\{}&{\mathrm{1}}&{{c}+\mathrm{1}}&{−{c}+\mathrm{1}}&{−{c}+\mathrm{5}}\end{pmatrix}\:\: \\ $$$$−{c}+\mathrm{5}=\mathrm{0}\Rightarrow{c}=\mathrm{5} \\ $$
Commented by Rio Michael last updated on 17/Jun/20

$$\mathrm{please}\:\mathrm{i}\:\mathrm{will}\:\mathrm{love}\:\mathrm{to}\:\mathrm{study}\:\mathrm{this}\:\mathrm{method}\:\mathrm{of} \\ $$$$\mathrm{yours}..\:\mathrm{please}\:\mathrm{hint}\:\mathrm{me}\:\mathrm{on}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{it}. \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jun/20

$${Let}\:{a}\:{polynomial}\:{P}\left({x}\right)\:{is}\:{divided} \\ $$$${by}\:{x}−{r}\:\left({Synthetically}\right). \\ $$$$\mathrm{1}.\:{Write}\:{P}\left({x}\right)\:{in}\:{descending}\:{order} \\ $$$${including}\:{missing}\:{terms}\:{with} \\ $$$${coefficients}\:\mathrm{0}.\:{For}\:{example} \\ $$$${if}\:{P}\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}+\mathrm{4}{x}^{\mathrm{3}} \\ $$$$\:{then}\:\:\:\:{P}\left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{0}{x}−\mathrm{8} \\ $$$$\mathrm{2}.{The}\:{divisor}\:{should}\:{be}\:{of}\:{type} \\ $$$${x}−{r}.\:{r}\:{is}\:{called}\:{multiplyer} \\ $$$${In}\:{x}−\mathrm{3},{the}\:{multiplyer}\:\mathrm{3}\:{and} \\ $$$${in}\:{x}+\mathrm{3}={x}−\left(−\mathrm{3}\right)\:{the}\:{multiplyer} \\ $$$${is}\:−\mathrm{3}. \\ $$$$\mathrm{3}.{First}\:{row}\:{consist}\:{of} \\ $$$${multiplyer},{separator}\:\&\:{coefficients}\:{of}\:{p}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Process} \\ $$$$\mathcal{E}{xample}. \\ $$$$\left({C}_{\mathrm{0}} {x}^{\mathrm{3}} +{C}_{\mathrm{1}} {x}^{\mathrm{2}} +{C}_{\mathrm{2}} {x}+{C}_{\mathrm{3}} \right)\boldsymbol{\div}\left({x}−{r}\right) \\ $$$$\left(\mathrm{1}\right),…,\left(\mathrm{7}\right)\:{are}\:{entries}\:{processed}\:{in}\:{order}. \\ $$$$\begin{pmatrix}{\left.{r}\right)}&{\:\:\:\:{C}_{\mathrm{0}} }&{{C}_{\mathrm{1}} }&{{C}_{\mathrm{2}} }&{{C}_{\mathrm{3}} }\\{}&{}&{\:\left(\mathrm{2}\right)}&{\left(\mathrm{4}\right)}&{\left(\mathrm{6}\right)}\\{}&{\:\left(\mathrm{1}\right)}&{\:\left(\mathrm{3}\right)}&{\left(\mathrm{5}\right)}&{\left(\mathrm{7}\right)}\end{pmatrix} \\ $$$$\left(\mathrm{1}\right)={C}_{\mathrm{0}} \\ $$$$\left(\mathrm{2}\right)=\left(\mathrm{1}\right).{r}={C}_{\mathrm{0}} .{r} \\ $$$$\left(\mathrm{3}\right)={C}_{\mathrm{1}} +\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)=\left(\mathrm{3}\right).{r} \\ $$$$\left(\mathrm{5}\right)={C}_{\mathrm{2}} +\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{6}\right)=\left(\mathrm{5}\right).{r} \\ $$$$\left(\mathrm{7}\right)={C}_{\mathrm{3}} +\left(\mathrm{6}\right)={remainder} \\ $$$$\left(\mathrm{1}\right){x}^{\mathrm{2}} +\left(\mathrm{3}\right){x}+\left(\mathrm{5}\right)={quotient} \\ $$
Commented by Rio Michael last updated on 17/Jun/20

$$\mathrm{wow},\:\mathrm{nice}\:\mathrm{method}\:\mathrm{sir}..\mathrm{thanks} \\ $$$$\mathrm{thanks}\:\mathrm{alot}\:\mathrm{sir}. \\ $$
Commented by bemath last updated on 18/Jun/20

$$\mathrm{Horner}\:\mathrm{method}\:\mathrm{sir} \\ $$
Answered by mathmax by abdo last updated on 16/Jun/20

$$\mathrm{p}\left(\mathrm{x}\right)\:=\mathrm{x}^{\mathrm{3}} −\mathrm{1}\:+\mathrm{cx}^{\mathrm{2}} −\mathrm{c}\:−\mathrm{2c}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{2c}\:+\mathrm{4}\:+\mathrm{1}+\mathrm{c} \\ $$$$=\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)+\mathrm{c}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)\:−\mathrm{2c}\left(\mathrm{x}−\mathrm{1}\right)\:+\mathrm{5}−\mathrm{c} \\ $$$$=\left(\mathrm{x}−\mathrm{1}\right)\left\{\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}\:+\mathrm{cx}−\mathrm{c}\right)\:+\mathrm{5}−\mathrm{c} \\ $$$$\mathrm{x}−\mathrm{1}\:\mathrm{divide}\:\mathrm{p}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{5}−\mathrm{c}\:=\mathrm{0}\:\Rightarrow\mathrm{c}=\mathrm{5}\:\mathrm{and}\:\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}−\mathrm{1}\right)\mathrm{Q}\left(\mathrm{x}\right)\:\mathrm{with} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{6x}−\mathrm{4} \\ $$