綜合除法200811071645

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屬性

已知：
1. 若 $f\left( x \right)=a\left( x-3 \right)^{4}+b\left( x-3 \right)^{3}+c\left( x-3 \right)^{2}+d\left( x-3 \right)+e$ ，求 $\left( a,b,c,d,e \right)$
2. 求 $f\left( 2.99 \right)$ 之近似值，至小數點後第二位，第三位以下四捨五入
3. 求 $f\left( 2+\sqrt{3}i \right)$

$f\left( x \right)=a\left( x-3 \right)^{4}+b\left( x-3 \right)^{3}+c\left( x-3 \right)^{2}+d\left( x-3 \right)+e$
$\Rightarrow \frac{f\left( x \right)}{x-3}=\frac{a\left( x-3 \right)^{4}+b\left( x-3 \right)^{3}+c\left( x-3 \right)^{2}+d\left( x-3 \right)+e}{x-3}$
$={\color{Blue}a\left( x-3 \right)^{3}+b\left( x-3 \right)^{2}+c\left( x-3 \right)^{1}+d}+\frac{{\color{Red}e}}{x-3}$
$\frac{{\color{Blue}a\left( x-3 \right)^{3}+b\left( x-3 \right)^{2}+c\left( x-3 \right)^{1}+d}}{x-3}={\color{Magenta}a\left( x-3 \right)^{2}+b\left( x-3 \right)^{1}+c}+\frac{{\color{Red}d}}{x-3}$
$\frac{{\color{Magenta}a\left( x-3 \right)^{2}+b\left( x-3 \right)^{1}+c}}{x-3}={\color{OliveGreen}a\left( x-3 \right)+b}+\frac{{\color{Red}c}}{x-3}$
$\frac{{\color{OliveGreen}a\left( x-3 \right)+b}}{x-3}={\color{Red}a}+\frac{{\color{Red}b}}{x-3}$
$\Rightarrow \left( a,b,c,d,e \right)=\left( 1,7,13,2,3 \right)\#$

$\left\{ \begin{align} & \left( a,b,c,d,e \right)=\left( 1,7,13,2,3 \right) \\ & f\left( x \right)=a\left( x-3 \right)^{4}+b\left( x-3 \right)^{3}+c\left( x-3 \right)^{2}+d\left( x-3 \right)+e \\ \end{align} \right.$
$\Rightarrow f\left( x \right)=\left( x-3 \right)^{4}+7\left( x-3 \right)^{3}+13\left( x-3 \right)^{2}+2\left( x-3 \right)+3$
$\Rightarrow f\left( 2.99 \right)={\color{Red}\left( -0.01 \right)^{4}+7\times \left( -0.01 \right)^{3}+13\times \left( -0.01 \right)^{2}}+2\times \left( -0.01 \right)+3$
$\approx 2\times \left( -0.01 \right)+3=-0.02+3=2.98\#$

已知： $f\left( x \right)=x^{4}-5x^{3}+4x^{2}+5x+6$
令： $\left\{ \begin{align} & g\left( x \right)=x^{2}+ax+b \\ & g\left( \alpha \right)=0,\alpha =2+\sqrt{3}i \\ & g\left( \beta \right)=0,\beta =2-\sqrt{3}i \\ \end{align} \right.$ ，則
$\left\{ \begin{align} & \alpha +\beta =-a=\left( 2+\sqrt{3}i \right)+\left( 2-\sqrt{3}i \right)=4 \\ & \alpha \times \beta =b=\left( 2+\sqrt{3}i \right)\times \left( 2-\sqrt{3}i \right)=2^{2}-\left( \sqrt{3}i \right)^{2}=4+3=7 \\ \end{align} \right.$
$\Rightarrow \left\{ \begin{align} & a=-4 \\ & b=7 \\ \end{align} \right.\Rightarrow g\left( x \right)=x^{2}-4x+7$
$\frac{f\left( x \right)}{g\left( x \right)}=\left( x^{2}-x-7 \right)+\frac{\left( -16x+55 \right)}{g\left( x \right)}$
$\Rightarrow f\left( x \right)=g\left( x \right)\times \left( x^{2}-x-7 \right)+\left( -16x+55 \right)$
$\Rightarrow f\left( 2+\sqrt{3}i \right)=f\left( \alpha \right)=g\left( \alpha \right)\times \left( \alpha ^{2}-\alpha -7 \right)+\left( -16\alpha +55 \right)=0\times \left( \alpha ^{2}-\alpha -7 \right)+\left( -16\alpha +55 \right)$
$=-16\alpha +55=-16\times \left( 2+\sqrt{3}i \right)+55=23-16\sqrt{3}i\#$