Divide $$$x^{2} \left(x - 3\right)$$$ by $$$x - 2$$$

Rewrite the dividend: $$$x^{2} \left(x - 3\right) = x^{3} - 3 x^{2}$$$.

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

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

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

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

Step 2

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

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

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

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

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

Step 3

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

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

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

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

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

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

The resulting table is shown once more:

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

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