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Question Number 135619 by liberty last updated on 14/Mar/21

Polynomial

$${Polynomial} \\ $$ 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?\\n

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

$$\mathrm{Use}\:\mathrm{Horner}\:\mathrm{Method}\: \\ $$ $$\mathrm{let}\:\mathrm{D}\left(\mathrm{x}\right)=\:\mathrm{x}^{\mathrm{3}} −\mathrm{9x}\:=\:\mathrm{0}\:,\:\mathrm{x}^{\mathrm{3}} =\mathrm{0}.\mathrm{x}^{\mathrm{2}} +\mathrm{9x}+\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\begin{array}{|c|c|c|c|c|c|}{\bullet}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{−\mathrm{8}}&\hline{\mathrm{1}}&\hline{−\mathrm{10}}&\hline{\mathrm{0}}&\hline{−\mathrm{12}}\\{\mathrm{0}}&\hline{\ast}&\hline{\mathrm{\color{mathred}{0}}}&\hline{\mathrm{\color{mathred}{9}}}&\hline{\mathrm{\color{mathred}{0}}}&\hline{}&\hline{}&\hline{}\\{\mathrm{9}}&\hline{\ast}&\hline{\ast}&\hline{\mathrm{\color{mathred}{0}}}&\hline{\mathrm{\color{mathred}{0}}}&\hline{\mathrm{\color{mathred}{0}}}&\hline{}&\hline{}\\{\mathrm{0}}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\mathrm{\color{mathred}{0}}}&\hline{\mathrm{\color{mathred}{9}}}&\hline{\mathrm{\color{mathred}{0}}}&\hline{}\\{\bullet}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\mathrm{\color{mathred}{0}}}&\hline{\mathrm{\color{mathred}{9}}}&\hline{\mathrm{\color{mathred}{0}}}\\{\bullet}&\hline{\mathrm{\color{mathred}{1}}}&\hline{\mathrm{\color{mathred}{0}}}&\hline{\mathrm{\color{mathred}{1}}}&\hline{\mathrm{\color{mathred}{1}}}&\hline{\color{mathred}{−}\mathrm{\color{mathred}{1}}}&\hline{\mathrm{\color{mathred}{9}}}&\hline{\color{mathred}{−}\mathrm{\color{mathred}{1}\color{mathred}{2}}}\\\hline\end{array} \\ $$ $$\mathrm{\color{mathred}{W}\color{mathred}{e}}\color{mathred}{\:}\mathrm{\color{mathred}{g}\color{mathred}{e}\color{mathred}{t}}\color{mathred}{\:}\mathrm{\color{mathred}{r}\color{mathred}{e}\color{mathred}{m}\color{mathred}{a}\color{mathred}{i}\color{mathred}{n}\color{mathred}{d}\color{mathred}{e}\color{mathred}{r}}\color{mathred}{\:}\mathrm{\color{mathred}{r}}\color{mathred}{\left(}\mathrm{\color{mathred}{x}}\color{mathred}{\right)}\color{mathred}{=}\color{mathblue}{−}\mathrm{\color{mathblue}{x}}^{\mathrm{\color{mathblue}{2}}} \color{mathblue}{+}\mathrm{\color{mathblue}{9}\color{mathblue}{x}}\color{mathblue}{−}\mathrm{\color{mathblue}{1}\color{mathblue}{2}} \\ $$ $$\mathrm{\color{mathblue}{a}\color{mathblue}{n}\color{mathblue}{d}}\color{mathblue}{\:}\mathrm{\color{mathblue}{q}\color{mathblue}{u}\color{mathblue}{o}\color{mathblue}{t}\color{mathblue}{i}\color{mathblue}{e}\color{mathblue}{n}\color{mathblue}{t}}\color{mathblue}{\:}\mathrm{\color{mathblue}{i}\color{mathblue}{s}}\color{mathblue}{\:}\mathrm{\color{mathblue}{q}}\color{mathblue}{\left(}\mathrm{\color{mathblue}{x}}\color{mathblue}{\right)}\color{mathblue}{=}\mathrm{\color{mathblue}{x}}^{\mathrm{\color{mathblue}{3}}} \color{mathblue}{+}\mathrm{\color{mathblue}{0}}\color{mathblue}{.}\mathrm{\color{mathblue}{x}}^{\mathrm{\color{mathblue}{2}}} \color{mathblue}{+}\mathrm{\color{mathblue}{x}}\color{mathblue}{+}\mathrm{\color{mathblue}{1}}\color{mathblue}{=}\mathrm{\color{mathblue}{x}}^{\mathrm{\color{mathblue}{3}}} \color{mathblue}{+}\mathrm{\color{mathblue}{x}}\color{mathblue}{+}\mathrm{\color{mathblue}{1}} \\ $$

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