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

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

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

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

Commented by imjagoll last updated on 04/Jul/21

$$\mathrm{yes}$$

Answered by liberty last updated on 04/Jul/21

$$by{Horner}^{\prime }stheorem$$$$d\left(x\right)=x^3+2x^2+8x+6=0$$$$\Rightarrow x^3=-2x^2-8x-6$$$$\overline{)\begin{array}{cccccc}\ast & 1& 4& 2p+2& 2p+5q+2& 3q+2r\\ -2& \ast & -2& -8& -6& \ast \\ -8& \ast & \ast & -4& -16& -12\\ -6& \ast & \ast & \ast & \ast & \ast \\ \ast & 1& 2& 2p-10& 2p+5q-20& 3q+2r-12\end{array}}$$$$weget\{\begin{array}{l}2p-10=0\to p=5\\ 10+5q-20=0\to q=2\\ 6+2r-12=0\to r=3\end{array}$$$$then(p+2q)r=(5+4)\times 3=27$$

Commented by imjagoll last updated on 04/Jul/21

$$\mathrm{yes}$$

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