将 $$$x^{3} + 7 x^{2} + 1$$$ 除以 $$$x - 1$$$
您的输入
使用长除法计算$$$\frac{x^{3} + 7 x^{2} + 1}{x - 1}$$$。
解答
将题目写成特殊格式(缺失项写为零系数):
$$$\begin{array}{r|r}\hline\\x-1&x^{3}+7 x^{2}+0 x+1\end{array}$$$
步骤 1
将被除式的首项除以除式的首项: $$$\frac{x^{3}}{x} = x^{2}$$$.
将计算结果写在表格的上部。
将其乘以除数:$$$x^{2} \left(x-1\right) = x^{3}- x^{2}$$$。
从得到的结果中减去被除数:$$$\left(x^{3}+7 x^{2}+1\right) - \left(x^{3}- x^{2}\right) = 8 x^{2}+1$$$
$$\begin{array}{r|rrrr:c}&{\color{GoldenRod}x^{2}}&&&&\\\hline\\{\color{Magenta}x}-1&{\color{GoldenRod}x^{3}}&+7 x^{2}&+0 x&+1&\frac{{\color{GoldenRod}x^{3}}}{{\color{Magenta}x}} = {\color{GoldenRod}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&{\color{GoldenRod}x^{2}} \left(x-1\right) = x^{3}- x^{2}\\\hline\\&&8 x^{2}&+0 x&+1&\end{array}$$步骤 2
将所得余式的首项除以除式的首项: $$$\frac{8 x^{2}}{x} = 8 x$$$
将计算结果写在表格的上部。
将其乘以除数:$$$8 x \left(x-1\right) = 8 x^{2}- 8 x$$$。
从得到的结果中减去余数:$$$\left(8 x^{2}+1\right) - \left(8 x^{2}- 8 x\right) = 8 x+1$$$
$$\begin{array}{r|rrrr:c}&x^{2}&{\color{Chartreuse}+8 x}&&&\\\hline\\{\color{Magenta}x}-1&x^{3}&+7 x^{2}&+0 x&+1&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&\\\hline\\&&{\color{Chartreuse}8 x^{2}}&+0 x&+1&\frac{{\color{Chartreuse}8 x^{2}}}{{\color{Magenta}x}} = {\color{Chartreuse}8 x}\\&&-\phantom{8 x^{2}}&&&\\&&8 x^{2}&- 8 x&&{\color{Chartreuse}8 x} \left(x-1\right) = 8 x^{2}- 8 x\\\hline\\&&&8 x&+1&\end{array}$$步骤 3
将所得余式的首项除以除式的首项: $$$\frac{8 x}{x} = 8$$$
将计算结果写在表格的上部。
将其乘以除数:$$$8 \left(x-1\right) = 8 x-8$$$。
从得到的结果中减去余数:$$$\left(8 x+1\right) - \left(8 x-8\right) = 9$$$
$$\begin{array}{r|rrrr:c}&x^{2}&+8 x&{\color{Brown}+8}&&\\\hline\\{\color{Magenta}x}-1&x^{3}&+7 x^{2}&+0 x&+1&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&\\\hline\\&&8 x^{2}&+0 x&+1&\\&&-\phantom{8 x^{2}}&&&\\&&8 x^{2}&- 8 x&&\\\hline\\&&&{\color{Brown}8 x}&+1&\frac{{\color{Brown}8 x}}{{\color{Magenta}x}} = {\color{Brown}8}\\&&&-\phantom{8 x}&&\\&&&8 x&-8&{\color{Brown}8} \left(x-1\right) = 8 x-8\\\hline\\&&&&9&\end{array}$$由于余式的次数小于除式的次数,故除法完成。
所得表格再次显示如下:
$$\begin{array}{r|rrrr:c}&{\color{GoldenRod}x^{2}}&{\color{Chartreuse}+8 x}&{\color{Brown}+8}&&\text{提示}\\\hline\\{\color{Magenta}x}-1&{\color{GoldenRod}x^{3}}&+7 x^{2}&+0 x&+1&\frac{{\color{GoldenRod}x^{3}}}{{\color{Magenta}x}} = {\color{GoldenRod}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&{\color{GoldenRod}x^{2}} \left(x-1\right) = x^{3}- x^{2}\\\hline\\&&{\color{Chartreuse}8 x^{2}}&+0 x&+1&\frac{{\color{Chartreuse}8 x^{2}}}{{\color{Magenta}x}} = {\color{Chartreuse}8 x}\\&&-\phantom{8 x^{2}}&&&\\&&8 x^{2}&- 8 x&&{\color{Chartreuse}8 x} \left(x-1\right) = 8 x^{2}- 8 x\\\hline\\&&&{\color{Brown}8 x}&+1&\frac{{\color{Brown}8 x}}{{\color{Magenta}x}} = {\color{Brown}8}\\&&&-\phantom{8 x}&&\\&&&8 x&-8&{\color{Brown}8} \left(x-1\right) = 8 x-8\\\hline\\&&&&9&\end{array}$$因此,$$$\frac{x^{3} + 7 x^{2} + 1}{x - 1} = \left(x^{2} + 8 x + 8\right) + \frac{9}{x - 1}$$$。
答案
$$$\frac{x^{3} + 7 x^{2} + 1}{x - 1} = \left(x^{2} + 8 x + 8\right) + \frac{9}{x - 1}$$$A