餘式定理200811131232

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屬性

$\left\{ \begin{align} & f\left( x \right)\text{ is a polynomial} \\ & \deg \left[ f\left( x \right) \right]\ge 3 \\ & R_{x^{2}+x+1}\left[ f\left( x \right) \right]=x+1 \\ & R_{x-2}\left[ f\left( x \right) \right]=10 \\ \end{align} \right.$

$R_{\left( x-2 \right)\left( x^{2}+x+1 \right)}\left[ f\left( x \right) \right]$

因： $R_{x^{2}+x+1}\left[ f\left( x \right) \right]=x+1$ ，
故令： $R_{\left( x-2 \right)\left( x^{2}+x+1 \right)}\left[ f\left( x \right) \right]=\left( x^{2}+x+1 \right)\times a+\left( x+1 \right)$ ，
故： $f\left( 2 \right)=\left( 2^{2}+2+1 \right)\times a+\left( 2+1 \right)=7a+3$
又： $R_{x-2}\left[ f\left( x \right) \right]=10\Rightarrow f\left( 2 \right)=10$
$\Rightarrow 7a+3=10\Rightarrow a=1$
所以： $R_{\left( x-2 \right)\left( x^{2}+x+1 \right)}\left[ f\left( x \right) \right]=\left( x^{2}+x+1 \right)\times 1+\left( x+1 \right)=x^{2}+2x+2\#$

令： $R_{\left( x-2 \right)\left( x^{2}+x+1 \right)}\left[ f\left( x \right) \right]=ax^{2}+bx+c$
$\Rightarrow f\left( x \right)=\left( x-2 \right)\left( x^{2}+x+1 \right)\times {q}'\left( x \right)+ax^{2}+bx+c$