Divide $$$x^{3} + 7 x^{2} + 1$$$ by $$$x - 1$$$

Write the problem in the special format (missed terms are written with zero coefficients):

$$$\begin{array}{r|r}\hline\\x-1&x^{3}+7 x^{2}+0 x+1\end{array}$$$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{x^{3}}{x} = x^{2}$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$x^{2} \left(x-1\right) = x^{3}- x^{2}$$$.

Subtract the dividend from the obtained result: $$$\left(x^{3}+7 x^{2}+1\right) - \left(x^{3}- x^{2}\right) = 8 x^{2}+1$$$.

$$\begin{array}{r|rrrr:c}&{\color{Green}x^{2}}&&&&\\\hline\\{\color{Magenta}x}-1&{\color{Green}x^{3}}&+7 x^{2}&+0 x&+1&\frac{{\color{Green}x^{3}}}{{\color{Magenta}x}} = {\color{Green}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&{\color{Green}x^{2}} \left(x-1\right) = x^{3}- x^{2}\\\hline\\&&8 x^{2}&+0 x&+1&\end{array}$$

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{8 x^{2}}{x} = 8 x$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$8 x \left(x-1\right) = 8 x^{2}- 8 x$$$.

Subtract the remainder from the obtained result: $$$\left(8 x^{2}+1\right) - \left(8 x^{2}- 8 x\right) = 8 x+1$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&{\color{Purple}+8 x}&&&\\\hline\\{\color{Magenta}x}-1&x^{3}&+7 x^{2}&+0 x&+1&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&\\\hline\\&&{\color{Purple}8 x^{2}}&+0 x&+1&\frac{{\color{Purple}8 x^{2}}}{{\color{Magenta}x}} = {\color{Purple}8 x}\\&&-\phantom{8 x^{2}}&&&\\&&8 x^{2}&- 8 x&&{\color{Purple}8 x} \left(x-1\right) = 8 x^{2}- 8 x\\\hline\\&&&8 x&+1&\end{array}$$

Step 3

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{8 x}{x} = 8$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$8 \left(x-1\right) = 8 x-8$$$.

Subtract the remainder from the obtained result: $$$\left(8 x+1\right) - \left(8 x-8\right) = 9$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&+8 x&{\color{OrangeRed}+8}&&\\\hline\\{\color{Magenta}x}-1&x^{3}&+7 x^{2}&+0 x&+1&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&\\\hline\\&&8 x^{2}&+0 x&+1&\\&&-\phantom{8 x^{2}}&&&\\&&8 x^{2}&- 8 x&&\\\hline\\&&&{\color{OrangeRed}8 x}&+1&\frac{{\color{OrangeRed}8 x}}{{\color{Magenta}x}} = {\color{OrangeRed}8}\\&&&-\phantom{8 x}&&\\&&&8 x&-8&{\color{OrangeRed}8} \left(x-1\right) = 8 x-8\\\hline\\&&&&9&\end{array}$$

Since the degree of the remainder is less than the degree of the divisor, we are done.

The resulting table is shown once more:

$$\begin{array}{r|rrrr:c}&{\color{Green}x^{2}}&{\color{Purple}+8 x}&{\color{OrangeRed}+8}&&\text{Hints}\\\hline\\{\color{Magenta}x}-1&{\color{Green}x^{3}}&+7 x^{2}&+0 x&+1&\frac{{\color{Green}x^{3}}}{{\color{Magenta}x}} = {\color{Green}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&- x^{2}&&&{\color{Green}x^{2}} \left(x-1\right) = x^{3}- x^{2}\\\hline\\&&{\color{Purple}8 x^{2}}&+0 x&+1&\frac{{\color{Purple}8 x^{2}}}{{\color{Magenta}x}} = {\color{Purple}8 x}\\&&-\phantom{8 x^{2}}&&&\\&&8 x^{2}&- 8 x&&{\color{Purple}8 x} \left(x-1\right) = 8 x^{2}- 8 x\\\hline\\&&&{\color{OrangeRed}8 x}&+1&\frac{{\color{OrangeRed}8 x}}{{\color{Magenta}x}} = {\color{OrangeRed}8}\\&&&-\phantom{8 x}&&\\&&&8 x&-8&{\color{OrangeRed}8} \left(x-1\right) = 8 x-8\\\hline\\&&&&9&\end{array}$$

Therefore, $$$\frac{x^{3} + 7 x^{2} + 1}{x - 1} = \left(x^{2} + 8 x + 8\right) + \frac{9}{x - 1}$$$.