筆算による整式の割り算

	計算過程	解説
ステップ1	$3$\displaystyle{x- 2\text{\hspace{1}}\bar{ \)\text{\hspace{1}}2{x}^{3}- 7{x}^{2}+7x+3 }}$	$3$\displaystyle{ 2{x}^{3}- 7{x}^{2}+7x+3 }$ は3次の整式で， $3$\displaystyle{ x- 2 }$ は1次の整式であるので，商の次数は2次になる．
ステップ2	$3$\displaystyle{\begin{array}{lll} & \text{\hspace{5}}\{\red{\text{there is tag which can't be converted to TeX}}\} & \vspace{6}\\ x- 2 & \bar{ \)\text{\hspace{1}}\{\red{\text{there is tag which can't be converted to TeX}}\}- 7{x}^{2}+7x+3 } & \vspace{6}\\ \end{array}}$	商の2次の項を定める． $3$\displaystyle{ 2{x}^{3} }$ の項を打ち消すので，商の2次の項は $3$2{x}^{2}$ となる．
ステップ3	$3$\displaystyle{\begin{array}{lll} & \text{\hspace{5}}2{x}^{2} & \vspace{6}\\ x- 2 & \bar{ \)\text{\hspace{1}}2{x}^{3}- 7{x}^{2}+7x+3 } & \vspace{6}\\ &\{\red{\text{there is tag which can't be converted to TeX}}\}& \vspace{6}\\ \end{array}}$	$3$\displaystyle{2{x}^{2}\times \left({ x- 2 }\right)}$ の計算をする．
ステップ4	$3$\displaystyle{\begin{array}{lll} & \text{\hspace{5}}2{x}^{2}\{\red{\text{there is tag which can't be converted to TeX}}\} & \vspace{6}\\ x- 2 & \bar{ \)\text{\hspace{1}}2{x}^{3}- 7{x}^{2}+7x+3 } & \vspace{6}\\ &\begin{array}{lll}\text{\hspace{1}}\text{\hspace{1}}\underline{ 2{x}^{3}- 4{x}^{2}\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}- 3{x}^{2}+7x& \vspace{6}\\ \end{array}& \vspace{6}\\ \end{array}}$	$3$\displaystyle{\left({ 2{x}^{3}- 7{x}^{2}+7x+3 }\right)- \left({ 2{x}^{3}- 4{x}^{2} }\right)}$ の計算をする．次に， $3$\displaystyle{ \text{\hspace{1}}- 3{x}^{2} }$ を打ち消すように商の1次の項を定める．
ステップ5	$3$\displaystyle{\begin{array}{lll} & \text{\hspace{5}}2{x}^{2}- 3x & \vspace{6}\\ x- 2 & \bar{ \)\text{\hspace{1}}2{x}^{3}- 7{x}^{2}+7x+3 } & \vspace{6}\\ &\begin{array}{lll}\text{\hspace{1}}\text{\hspace{1}}\underline{ 2{x}^{3}- 4{x}^{2}\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}- 3{x}^{2}+7x& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}- 3{x}^{2}+6x& \vspace{6}\\ \end{array}& \vspace{6}\\ \end{array}}$	$3$\displaystyle{\text{\hspace{1}}- 3{x}^{2}\times \left({ x- 2 }\right)}$ の計算をする．
ステップ6	$3$\displaystyle{\begin{array}{lll} & \text{\hspace{5}}2{x}^{2}- 3x+1 & \vspace{6}\\ x- 2 & \bar{ \)\text{\hspace{1}}2{x}^{3}- 7{x}^{2}+7x+3 } & \vspace{6}\\ &\begin{array}{lll}\text{\hspace{1}}\text{\hspace{1}}\underline{ 2{x}^{3}- 4{x}^{2}\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}- 3{x}^{2}+7x& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\underline{ - 3{x}^{2}+6x\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\text{\hspace{1}}x+3& \vspace{6}\\ \end{array}& \vspace{6}\\ \end{array}}$	$3$\displaystyle{\left({ - 3{x}^{2}+7x }\right)- \left({ - 3{x}^{2}+6x }\right)}$ の計算をする．次に， $3$\displaystyle{x}$ を打ち消すために商の定数項を定める．
ステップ7	$3$\displaystyle{\begin{array}{lll} & \text{\hspace{5}}2{x}^{2}- 3x+1 & \vspace{6}\\ x- 2 & \bar{ \)\text{\hspace{1}}2{x}^{3}- 7{x}^{2}+7x+3 } & \vspace{6}\\ &\begin{array}{lll}\text{\hspace{1}}\text{\hspace{1}}\underline{ 2{x}^{3}- 4{x}^{2}\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}- 3{x}^{2}+7x& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\underline{ - 3{x}^{2}+6x\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\text{\hspace{1}}x+3& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\text{\hspace{1}}x- 2& \vspace{6}\\ \end{array}& \vspace{6}\\ \end{array}}$	$3$\displaystyle{1\times \left({ x- 2 }\right)}$ の計算をする．
ステップ8	$3$\displaystyle{\begin{array}{lll} & \text{\hspace{5}}2{x}^{2}- 3x+1 & \vspace{6}\\ x- 2 & \bar{ \)\text{\hspace{1}}2{x}^{3}- 7{x}^{2}+7x+3 } & \vspace{6}\\ &\begin{array}{lll}\text{\hspace{1}}\text{\hspace{1}}\underline{ 2{x}^{3}- 4{x}^{2}\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}- 3{x}^{2}+7x& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\underline{ - 3{x}^{2}+6x\text{\hspace{5}}\text{\hspace{5}} }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\text{\hspace{1}}x+3& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\text{\hspace{1}}\underline{ x- 2 }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{1}}\text{\hspace{1}}\text{\hspace{5}}\text{\hspace{1}}\text{\hspace{1}}5& \vspace{6}\\ \end{array}& \vspace{6}\\ \end{array}}$	$3$\displaystyle{\left({ x+3 }\right)- \left({ x- 2 }\right)}$ の計算をする．商が $3$\displaystyle{ 2{x}^{2}- 3x+1 }$ ，余りが $3$\displaystyle{5}$ となる．

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最終更新日： 2024年6月26日