Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

p(x)=q(x)(x−1) p(1)=0 1+c−2c+4=0 ⇒c=5

$${p}\left({x}\right)={q}\left({x}\right)\left({x}−\mathrm{1}\right) \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{1}+{c}−\mathrm{2}{c}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{c}=\mathrm{5} \\ $$

Answered by Rasheed.Sindhi last updated on 16/Jun/20

(((1)),1,( c),(−2c),( 4)),(,,( 1),( c+1),(−c+1)),(,1,(c+1),(−c+1),(−c+5)) ) −c+5=0⇒c=5

$$\begin{pmatrix}{\left.\mathrm{1}\right)}&{\mathrm{1}}&{\:\:\:{c}}&{−\mathrm{2}{c}}&{\:\:\:\:\:\mathrm{4}}\\{}&{}&{\:\:\:\mathrm{1}}&{\:\:{c}+\mathrm{1}}&{−{c}+\mathrm{1}}\\{}&{\mathrm{1}}&{{c}+\mathrm{1}}&{−{c}+\mathrm{1}}&{−{c}+\mathrm{5}}\end{pmatrix}\:\: \\ $$$$−{c}+\mathrm{5}=\mathrm{0}\Rightarrow{c}=\mathrm{5} \\ $$

Commented by Rio Michael last updated on 17/Jun/20

please i will love to study this method of yours.. please hint me on how to use it.

$$\mathrm{please}\:\mathrm{i}\:\mathrm{will}\:\mathrm{love}\:\mathrm{to}\:\mathrm{study}\:\mathrm{this}\:\mathrm{method}\:\mathrm{of} \\ $$$$\mathrm{yours}..\:\mathrm{please}\:\mathrm{hint}\:\mathrm{me}\:\mathrm{on}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{it}. \\ $$

Commented by Rasheed.Sindhi last updated on 17/Jun/20

Let a polynomial P(x) is divided by x−r (Synthetically). 1. Write P(x) in descending order including missing terms with coefficients 0. For example if P(x)=3x^2 −8+4x^3 then P(x)=4x^3 +3x^2 +0x−8 2.The divisor should be of type x−r. r is called multiplyer In x−3,the multiplyer 3 and in x+3=x−(−3) the multiplyer is −3. 3.First row consist of multiplyer,separator & coefficients of p(x) Process Example. (C_0 x^3 +C_1 x^2 +C_2 x+C_3 )÷(x−r) (1),...,(7) are entries processed in order. (((r)),( C_0 ),C_1 ,C_2 ,C_3 ),(,,( (2)),((4)),((6))),(,( (1)),( (3)),((5)),((7))) ) (1)=C_0 (2)=(1).r=C_0 .r (3)=C_1 +(2) (4)=(3).r (5)=C_2 +(4) (6)=(5).r (7)=C_3 +(6)=remainder (1)x^2 +(3)x+(5)=quotient

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

Commented by Rio Michael last updated on 17/Jun/20

wow, nice method sir..thanks thanks alot sir.

$$\mathrm{wow},\:\mathrm{nice}\:\mathrm{method}\:\mathrm{sir}..\mathrm{thanks} \\ $$$$\mathrm{thanks}\:\mathrm{alot}\:\mathrm{sir}. \\ $$

Commented by bemath last updated on 18/Jun/20

Horner method sir

$$\mathrm{Horner}\:\mathrm{method}\:\mathrm{sir} \\ $$

Answered by mathmax by abdo last updated on 16/Jun/20

p(x) =x^3 −1 +cx^2 −c −2c(x−1)−2c +4 +1+c =(x−1)(x^2 +x+1)+c(x−1)(x+1) −2c(x−1) +5−c =(x−1){x^2 +x+1 +cx−c) +5−c x−1 divide p(x) ⇒5−c =0 ⇒c=5 and p(x)=(x−1)Q(x) with Q(x)=x^2 +6x−4

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com