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Polynomial-Given-the-polynomial-P-x-x-6-8x-4-x-3-10x-2-12-divided-by-x-x-3-x-3-What-is-the-remainder-Is-there-a-formula-for-this-




Question Number 135619 by liberty last updated on 14/Mar/21
Polynomial  Given the polynomial P(x) = x^6 - 8x^4 + x^3 - 10x^2 - 12 divided by x(x-3) (x+3). What is the remainder? Is there a formula for this?
$${Polynomial} \\ $$Given the polynomial P(x) = x^6 – 8x^4 + x^3 – 10x^2 – 12 divided by x(x-3) (x+3). What is the remainder? Is there a formula for this?
Answered by EDWIN88 last updated on 14/Mar/21
Use Horner Method   let D(x)= x^3 −9x = 0 , x^3 =0.x^2 +9x+0          determinant ((•,1,0,(−8),1,(−10),0,(−12)),(0,∗,0,9,0,,,),(9,∗,∗,0,0,0,,),(0,∗,∗,∗,0,9,0,),(•,∗,∗,∗,∗,0,9,0),(•,1,0,1,1,(−1),9,(−12)))  We get remainder r(x)=−x^2 +9x−12  and quotient is q(x)=x^3 +0.x^2 +x+1=x^3 +x+1
$$\mathrm{Use}\:\mathrm{Horner}\:\mathrm{Method}\: \\ $$$$\mathrm{let}\:\mathrm{D}\left(\mathrm{x}\right)=\:\mathrm{x}^{\mathrm{3}} −\mathrm{9x}\:=\:\mathrm{0}\:,\:\mathrm{x}^{\mathrm{3}} =\mathrm{0}.\mathrm{x}^{\mathrm{2}} +\mathrm{9x}+\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\begin{array}{|c|c|c|c|c|c|}{\bullet}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{−\mathrm{8}}&\hline{\mathrm{1}}&\hline{−\mathrm{10}}&\hline{\mathrm{0}}&\hline{−\mathrm{12}}\\{\mathrm{0}}&\hline{\ast}&\hline{\mathrm{0}}&\hline{\mathrm{9}}&\hline{\mathrm{0}}&\hline{}&\hline{}&\hline{}\\{\mathrm{9}}&\hline{\ast}&\hline{\ast}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{}&\hline{}\\{\mathrm{0}}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\mathrm{0}}&\hline{\mathrm{9}}&\hline{\mathrm{0}}&\hline{}\\{\bullet}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\mathrm{0}}&\hline{\mathrm{9}}&\hline{\mathrm{0}}\\{\bullet}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{−\mathrm{1}}&\hline{\mathrm{9}}&\hline{−\mathrm{12}}\\\hline\end{array} \\ $$$$\mathrm{We}\:\mathrm{get}\:\mathrm{remainder}\:\mathrm{r}\left(\mathrm{x}\right)=−\mathrm{x}^{\mathrm{2}} +\mathrm{9x}−\mathrm{12} \\ $$$$\mathrm{and}\:\mathrm{quotient}\:\mathrm{is}\:\mathrm{q}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{0}.\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1} \\ $$

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