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Question-200336




Question Number 200336 by faysal last updated on 17/Nov/23
Answered by cortano12 last updated on 17/Nov/23
   determinant ((,1,2,(−3),5,(−2),3),((x=2),∗,2,8,(10),(30),(56)),(,1,4,5,(15),(28),(59)))   the remainder is 59  the quotient is h(x)=x^4 +4x^3 +5x^2 +15x+28
$$\:\:\begin{array}{|c|c|c|}{}&\hline{\mathrm{1}}&\hline{\mathrm{2}}&\hline{−\mathrm{3}}&\hline{\mathrm{5}}&\hline{−\mathrm{2}}&\hline{\mathrm{3}}\\{\mathrm{x}=\mathrm{2}}&\hline{\ast}&\hline{\mathrm{2}}&\hline{\mathrm{8}}&\hline{\mathrm{10}}&\hline{\mathrm{30}}&\hline{\mathrm{56}}\\{}&\hline{\mathrm{1}}&\hline{\mathrm{4}}&\hline{\mathrm{5}}&\hline{\mathrm{15}}&\hline{\mathrm{28}}&\hline{\mathrm{59}}\\\hline\end{array} \\ $$$$\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{59} \\ $$$$\mathrm{the}\:\mathrm{quotient}\:\mathrm{is}\:\mathrm{h}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{3}} +\mathrm{5x}^{\mathrm{2}} +\mathrm{15x}+\mathrm{28} \\ $$

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