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Question Number 144736 by mathdanisur last updated on 28/Jun/21

$$P\left(z\right)={az}^3+z^2-(a+6)z+b-6$$$$P\left(z\right)=(z^2+4)\cdot Q\left(z\right)$$$$Finda\cdot b=?$$

Answered by nimnim last updated on 12/Mar/22

$$\mathrm{P}\left(\mathrm{z}\right)={\mathrm{az}}^3+{\mathrm{z}}^2-(\mathrm{a}+6)\mathrm{z}+\mathrm{b}-6$$$$\mathrm{P}\left(\mathrm{z}\right)=({\mathrm{z}}^2+4)\mathrm{Q}\left(\mathrm{z}\right)$$$$=({\mathrm{z}}^2+4)(\mathrm{az}+\mathrm{k})$$$$={\mathrm{az}}^3+{\mathrm{kz}}^2+4\mathrm{az}+4\mathrm{k}$$$$\Rightarrow \mathrm{k}=1,4\mathrm{a}=-(\mathrm{a}+6),4\mathrm{k}=\mathrm{b}-6$$$$\Rightarrow \mathrm{k}=1,\mathrm{a}=-\frac{6}{5},\mathrm{b}=10$$$$\therefore \mathrm{a}.\mathrm{b}=-\frac{6}{5}.10=-12$$

Commented by mathdanisur last updated on 28/Jun/21

$$coolthanksSir$$

Answered by imjagoll last updated on 28/Jun/21

$$\overline{)\begin{array}{ccccc}\bullet & \mathrm{a}& 1& -\mathrm{a}-6& \mathrm{b}-6\\ 0& \ast & 0& -4\mathrm{a}& \ast \\ -4& \ast & \ast & 0& -4\\ \ast & \mathrm{a}& 1& -5\mathrm{a}-6& \mathrm{b}-10\end{array}}$$$$\iff -5\mathrm{a}-6=0;\mathrm{a}=-\frac{6}{5}$$$$\Rightarrow \mathrm{b}-10=0;\mathrm{b}=10$$$$\mathrm{therefore}\mathrm{a}\times \mathrm{b}=-12$$

Commented by mathdanisur last updated on 28/Jun/21

$$coolsolutionthanksSir$$

Commented by Rasheed.Sindhi last updated on 28/Jun/21

$$imjagollsir,I^{\prime }minterestedin$$$$yourmethod.Pleaseclerifyhow$$$$haveyoumadethetable?$$

Commented by liberty last updated on 29/Jun/21

$$\mathrm{i}\mathrm{think}\mathrm{this}\mathrm{is}\mathrm{by}{\mathrm{Horner}}^{\prime }\mathrm{s}\mathrm{Theorem}$$

Commented by imjagoll last updated on 29/Jun/21

$$\mathrm{yes}$$

Commented by Rasheed.Sindhi last updated on 29/Jun/21

$$\mathcal{T}h\alpha n\mathcal{X}bothsirs!$$

Answered by Rasheed.Sindhi last updated on 28/Jun/21

$$P\left(z\right)={az}^3+z^2-(a+6)z+b-6=(z^2+4)\cdot Q\left(z\right)$$$$Letz=2i(Inordertomake{z}^2+4zero)$$$$p\left(2i\right)=a{\left(2i\right)}^3+{\left(2i\right)}^2-(a+6)\left(2i\right)+b-6=0$$$$a.8i^2.i+4i^2-2(a+6)i+b-6=0$$$$-8ai-4-2(a+6)i+b-6=0$$$$(-8a-2a-12)i+b-10=0$$$$-10a-12=0\wedge b-10=0$$$$a=-12/10=-\frac{6}{5}\wedge b=10$$$$ab=(-\frac{6}{5})\left(10\right)=-12$$

Commented by mathdanisur last updated on 28/Jun/21

$$coolsolutionSirthanks$$

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