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Question Number 145294 by imjagoll last updated on 04/Jul/21

Given a polynomial p(x)=x^4 +4x^3 +(2p+2)x^2 +(2p+5q+2)x+3q+2r. If p(x)= (x^3 +2x^2 +8x+6)Q(x) then what the value of (p+2q)r .

$$\mathrm{Given}\:\mathrm{a}\:\mathrm{polynomial}\: \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{3}} +\left(\mathrm{2p}+\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{2p}+\mathrm{5q}+\mathrm{2}\right)\mathrm{x}+\mathrm{3q}+\mathrm{2r}. \\ $$$$\mathrm{If}\:\mathrm{p}\left(\mathrm{x}\right)=\:\left(\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{6}\right)\mathrm{Q}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{then}\:\mathrm{what}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\left(\mathrm{p}+\mathrm{2q}\right)\mathrm{r}\:. \\ $$

Answered by Rasheed.Sindhi last updated on 04/Jul/21

p(x)=x^4 +4x^3 +(2p+2)x^2 +(2p+5q+2)x+3q+2r. If p(x)= (x^3 +2x^2 +8x+6)Q(x) (p+2q)r =? _(−) Let Q(x)=x+k Given: D(x)=x^3 +2x^2 +8x+6 We can also consider: D(x)=x+k & Q(x)=x^3 +2x^2 +8x+6 By synthetic division : determinant (((−k)),1,( 4),(2p+2),(2p+5q+2),(3q+2r)),(,,(−k),( ...),( ...),( ...)),(,1,( 2),( 8),( 6),( 0))) 4−k=2⇒k=2 determinant (((−2)),1,( 4),(2p+2),(2p+5q+2),(3q+2r)),(,,(−2),( −4),( −16),( −12)),(,1,( 2),( 8),( 6),( 0))) 2p+2−4=8⇒p=5 2p+5q+2−16=6 2(5)+5q=20⇒q=2 3q+2r−12=0 3(2)+2r=12⇒r=3 (p+2q)r =( 5+2(2) )(3)=27

$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{3}} +\left(\mathrm{2p}+\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{2p}+\mathrm{5q}+\mathrm{2}\right)\mathrm{x}+\mathrm{3q}+\mathrm{2r}. \\ $$$$\mathrm{If}\:\mathrm{p}\left(\mathrm{x}\right)=\:\left(\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{6}\right)\mathrm{Q}\left(\mathrm{x}\right) \\ $$$$\underset{−} {\:\left(\mathrm{p}+\mathrm{2q}\right)\mathrm{r}\:=?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} \\ $$$$\mathrm{Let}\:\mathrm{Q}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{k} \\ $$$$\mathrm{Given}:\:\mathrm{D}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{6} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{also}\:\mathrm{consider}: \\ $$$$\mathrm{D}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{k}\:\:\&\:\:\mathrm{Q}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{6} \\ $$$$\mathcal{B}{y}\:{synthetic}\:{division}\:: \\ $$$$\begin{array}{|c|c|c|}{\left.−\mathrm{k}\right)}&\hline{\mathrm{1}}&\hline{\:\:\:\:\mathrm{4}}&\hline{\mathrm{2p}+\mathrm{2}}&\hline{\mathrm{2p}+\mathrm{5q}+\mathrm{2}}&\hline{\mathrm{3q}+\mathrm{2r}}\\{}&\hline{}&\hline{−\mathrm{k}}&\hline{\:\:\:\:...}&\hline{\:\:\:\:\:\:\:...}&\hline{\:\:\:...}\\{}&\hline{\mathrm{1}}&\hline{\:\:\:\mathrm{2}}&\hline{\:\:\:\mathrm{8}}&\hline{\:\:\:\:\:\:\:\mathrm{6}}&\hline{\:\:\:\:\mathrm{0}}\\\hline\end{array}\: \\ $$$$\mathrm{4}−\mathrm{k}=\mathrm{2}\Rightarrow\mathrm{k}=\mathrm{2} \\ $$$$\begin{array}{|c|c|c|}{\left.−\mathrm{2}\right)}&\hline{\mathrm{1}}&\hline{\:\:\:\:\mathrm{4}}&\hline{\mathrm{2p}+\mathrm{2}}&\hline{\mathrm{2p}+\mathrm{5q}+\mathrm{2}}&\hline{\mathrm{3q}+\mathrm{2r}}\\{}&\hline{}&\hline{−\mathrm{2}}&\hline{\:\:−\mathrm{4}}&\hline{\:\:\:\:\:−\mathrm{16}}&\hline{\:−\mathrm{12}}\\{}&\hline{\mathrm{1}}&\hline{\:\:\mathrm{2}}&\hline{\:\:\:\:\mathrm{8}}&\hline{\:\:\:\:\:\:\:\:\mathrm{6}}&\hline{\:\:\:\:\mathrm{0}}\\\hline\end{array}\: \\ $$$$\mathrm{2p}+\mathrm{2}−\mathrm{4}=\mathrm{8}\Rightarrow\mathrm{p}=\mathrm{5} \\ $$$$\mathrm{2p}+\mathrm{5q}+\mathrm{2}−\mathrm{16}=\mathrm{6} \\ $$$$\mathrm{2}\left(\mathrm{5}\right)+\mathrm{5q}=\mathrm{20}\Rightarrow\mathrm{q}=\mathrm{2} \\ $$$$\mathrm{3q}+\mathrm{2r}−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{3}\left(\mathrm{2}\right)+\mathrm{2r}=\mathrm{12}\Rightarrow\mathrm{r}=\mathrm{3} \\ $$$$\left(\mathrm{p}+\mathrm{2q}\right)\mathrm{r}\:=\left(\:\mathrm{5}+\mathrm{2}\left(\mathrm{2}\right)\:\right)\left(\mathrm{3}\right)=\mathrm{27} \\ $$

Commented by imjagoll last updated on 04/Jul/21

yes

$$\mathrm{yes} \\ $$

Answered by liberty last updated on 04/Jul/21

by Horner′s theorem d(x)=x^3 +2x^2 +8x+6=0 ⇒x^3 =−2x^2 −8x−6 determinant ((∗,1,4,(2p+2),(2p+5q+2),(3q+2r)),((−2),∗,(−2),(−8),(−6),∗),((−8),∗,∗,(−4),(−16),(−12)),((−6),∗,∗,∗,∗,∗),(∗,1,2,(2p−10),(2p+5q−20),(3q+2r−12))) we get { ((2p−10=0→p=5)),((10+5q−20=0→q=2)),((6+2r−12=0→r=3)) :} then (p+2q)r=(5+4)×3=27

$${by}\:{Horner}'{s}\:{theorem} \\ $$$${d}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{6} \\ $$$$\:\begin{array}{|c|c|c|c|c|}{\ast}&\hline{\mathrm{1}}&\hline{\mathrm{4}}&\hline{\mathrm{2}{p}+\mathrm{2}}&\hline{\mathrm{2}{p}+\mathrm{5}{q}+\mathrm{2}}&\hline{\mathrm{3}{q}+\mathrm{2}{r}}\\{−\mathrm{2}}&\hline{\ast}&\hline{−\mathrm{2}}&\hline{−\mathrm{8}}&\hline{−\mathrm{6}}&\hline{\ast}\\{−\mathrm{8}}&\hline{\ast}&\hline{\ast}&\hline{−\mathrm{4}}&\hline{−\mathrm{16}}&\hline{−\mathrm{12}}\\{−\mathrm{6}}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\ast}\\{\ast}&\hline{\mathrm{1}}&\hline{\mathrm{2}}&\hline{\mathrm{2}{p}−\mathrm{10}}&\hline{\mathrm{2}{p}+\mathrm{5}{q}−\mathrm{20}}&\hline{\mathrm{3}{q}+\mathrm{2}{r}−\mathrm{12}}\\\hline\end{array} \\ $$$$\:{we}\:{get}\:\begin{cases}{\mathrm{2}{p}−\mathrm{10}=\mathrm{0}\rightarrow{p}=\mathrm{5}}\\{\mathrm{10}+\mathrm{5}{q}−\mathrm{20}=\mathrm{0}\rightarrow{q}=\mathrm{2}}\\{\mathrm{6}+\mathrm{2}{r}−\mathrm{12}=\mathrm{0}\rightarrow{r}=\mathrm{3}}\end{cases} \\ $$$${then}\:\left({p}+\mathrm{2}{q}\right){r}=\left(\mathrm{5}+\mathrm{4}\right)×\mathrm{3}=\mathrm{27} \\ $$

Commented by imjagoll last updated on 04/Jul/21

yes

$$\mathrm{yes} \\ $$

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