Divide $$$x^{3}$$$ by $$$x + 2$$$

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

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

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: $$$x^{2} \left(x+2\right) = x^{3}+2 x^{2}$$$.

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

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

Step 2

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

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

Multiply it by the divisor: $$$- 2 x \left(x+2\right) = - 2 x^{2}- 4 x$$$.

Subtract the remainder from the obtained result: $$$\left(- 2 x^{2}\right) - \left(- 2 x^{2}- 4 x\right) = 4 x$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&{\color{DarkBlue}- 2 x}&&&\\\hline\\{\color{Magenta}x}+2&x^{3}&+0 x^{2}&+0 x&+0&\\&-\phantom{x^{3}}&&&&\\&x^{3}&+2 x^{2}&&&\\\hline\\&&{\color{DarkBlue}- 2 x^{2}}&+0 x&+0&\frac{{\color{DarkBlue}- 2 x^{2}}}{{\color{Magenta}x}} = {\color{DarkBlue}- 2 x}\\&&-\phantom{- 2 x^{2}}&&&\\&&- 2 x^{2}&- 4 x&&{\color{DarkBlue}- 2 x} \left(x+2\right) = - 2 x^{2}- 4 x\\\hline\\&&&4 x&+0&\end{array}$$

Step 3

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{4 x}{x} = 4$$$.

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

Multiply it by the divisor: $$$4 \left(x+2\right) = 4 x+8$$$.

Subtract the remainder from the obtained result: $$$\left(4 x\right) - \left(4 x+8\right) = -8$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&- 2 x&{\color{SaddleBrown}+4}&&\\\hline\\{\color{Magenta}x}+2&x^{3}&+0 x^{2}&+0 x&+0&\\&-\phantom{x^{3}}&&&&\\&x^{3}&+2 x^{2}&&&\\\hline\\&&- 2 x^{2}&+0 x&+0&\\&&-\phantom{- 2 x^{2}}&&&\\&&- 2 x^{2}&- 4 x&&\\\hline\\&&&{\color{SaddleBrown}4 x}&+0&\frac{{\color{SaddleBrown}4 x}}{{\color{Magenta}x}} = {\color{SaddleBrown}4}\\&&&-\phantom{4 x}&&\\&&&4 x&+8&{\color{SaddleBrown}4} \left(x+2\right) = 4 x+8\\\hline\\&&&&-8&\end{array}$$

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

The resulting table is shown once more:

$$\begin{array}{r|rrrr:c}&{\color{Crimson}x^{2}}&{\color{DarkBlue}- 2 x}&{\color{SaddleBrown}+4}&&\text{Hints}\\\hline\\{\color{Magenta}x}+2&{\color{Crimson}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{Crimson}x^{3}}}{{\color{Magenta}x}} = {\color{Crimson}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&+2 x^{2}&&&{\color{Crimson}x^{2}} \left(x+2\right) = x^{3}+2 x^{2}\\\hline\\&&{\color{DarkBlue}- 2 x^{2}}&+0 x&+0&\frac{{\color{DarkBlue}- 2 x^{2}}}{{\color{Magenta}x}} = {\color{DarkBlue}- 2 x}\\&&-\phantom{- 2 x^{2}}&&&\\&&- 2 x^{2}&- 4 x&&{\color{DarkBlue}- 2 x} \left(x+2\right) = - 2 x^{2}- 4 x\\\hline\\&&&{\color{SaddleBrown}4 x}&+0&\frac{{\color{SaddleBrown}4 x}}{{\color{Magenta}x}} = {\color{SaddleBrown}4}\\&&&-\phantom{4 x}&&\\&&&4 x&+8&{\color{SaddleBrown}4} \left(x+2\right) = 4 x+8\\\hline\\&&&&-8&\end{array}$$

Therefore, $$$\frac{x^{3}}{x + 2} = \left(x^{2} - 2 x + 4\right) + \frac{-8}{x + 2}$$$.