Divide $$$2 x^{3} - x^{2} - 12$$$ by $$$x + 3$$$

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

$$$\begin{array}{r|r}\hline\\x+3&2 x^{3}- x^{2}+0 x-12\end{array}$$$

Step 1

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

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

Multiply it by the divisor: $$$2 x^{2} \left(x+3\right) = 2 x^{3}+6 x^{2}$$$.

Subtract the dividend from the obtained result: $$$\left(2 x^{3}- x^{2}-12\right) - \left(2 x^{3}+6 x^{2}\right) = - 7 x^{2}-12$$$.

$$\begin{array}{r|rrrr:c}&{\color{Chocolate}2 x^{2}}&&&&\\\hline\\{\color{Magenta}x}+3&{\color{Chocolate}2 x^{3}}&- x^{2}&+0 x&-12&\frac{{\color{Chocolate}2 x^{3}}}{{\color{Magenta}x}} = {\color{Chocolate}2 x^{2}}\\&-\phantom{2 x^{3}}&&&&\\&2 x^{3}&+6 x^{2}&&&{\color{Chocolate}2 x^{2}} \left(x+3\right) = 2 x^{3}+6 x^{2}\\\hline\\&&- 7 x^{2}&+0 x&-12&\end{array}$$

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{- 7 x^{2}}{x} = - 7 x$$$.

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

Multiply it by the divisor: $$$- 7 x \left(x+3\right) = - 7 x^{2}- 21 x$$$.

Subtract the remainder from the obtained result: $$$\left(- 7 x^{2}-12\right) - \left(- 7 x^{2}- 21 x\right) = 21 x-12$$$.

$$\begin{array}{r|rrrr:c}&2 x^{2}&{\color{BlueViolet}- 7 x}&&&\\\hline\\{\color{Magenta}x}+3&2 x^{3}&- x^{2}&+0 x&-12&\\&-\phantom{2 x^{3}}&&&&\\&2 x^{3}&+6 x^{2}&&&\\\hline\\&&{\color{BlueViolet}- 7 x^{2}}&+0 x&-12&\frac{{\color{BlueViolet}- 7 x^{2}}}{{\color{Magenta}x}} = {\color{BlueViolet}- 7 x}\\&&-\phantom{- 7 x^{2}}&&&\\&&- 7 x^{2}&- 21 x&&{\color{BlueViolet}- 7 x} \left(x+3\right) = - 7 x^{2}- 21 x\\\hline\\&&&21 x&-12&\end{array}$$

Step 3

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{21 x}{x} = 21$$$.

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

Multiply it by the divisor: $$$21 \left(x+3\right) = 21 x+63$$$.

Subtract the remainder from the obtained result: $$$\left(21 x-12\right) - \left(21 x+63\right) = -75$$$.

$$\begin{array}{r|rrrr:c}&2 x^{2}&- 7 x&{\color{Green}+21}&&\\\hline\\{\color{Magenta}x}+3&2 x^{3}&- x^{2}&+0 x&-12&\\&-\phantom{2 x^{3}}&&&&\\&2 x^{3}&+6 x^{2}&&&\\\hline\\&&- 7 x^{2}&+0 x&-12&\\&&-\phantom{- 7 x^{2}}&&&\\&&- 7 x^{2}&- 21 x&&\\\hline\\&&&{\color{Green}21 x}&-12&\frac{{\color{Green}21 x}}{{\color{Magenta}x}} = {\color{Green}21}\\&&&-\phantom{21 x}&&\\&&&21 x&+63&{\color{Green}21} \left(x+3\right) = 21 x+63\\\hline\\&&&&-75&\end{array}$$

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

The resulting table is shown once more:

$$\begin{array}{r|rrrr:c}&{\color{Chocolate}2 x^{2}}&{\color{BlueViolet}- 7 x}&{\color{Green}+21}&&\text{Hints}\\\hline\\{\color{Magenta}x}+3&{\color{Chocolate}2 x^{3}}&- x^{2}&+0 x&-12&\frac{{\color{Chocolate}2 x^{3}}}{{\color{Magenta}x}} = {\color{Chocolate}2 x^{2}}\\&-\phantom{2 x^{3}}&&&&\\&2 x^{3}&+6 x^{2}&&&{\color{Chocolate}2 x^{2}} \left(x+3\right) = 2 x^{3}+6 x^{2}\\\hline\\&&{\color{BlueViolet}- 7 x^{2}}&+0 x&-12&\frac{{\color{BlueViolet}- 7 x^{2}}}{{\color{Magenta}x}} = {\color{BlueViolet}- 7 x}\\&&-\phantom{- 7 x^{2}}&&&\\&&- 7 x^{2}&- 21 x&&{\color{BlueViolet}- 7 x} \left(x+3\right) = - 7 x^{2}- 21 x\\\hline\\&&&{\color{Green}21 x}&-12&\frac{{\color{Green}21 x}}{{\color{Magenta}x}} = {\color{Green}21}\\&&&-\phantom{21 x}&&\\&&&21 x&+63&{\color{Green}21} \left(x+3\right) = 21 x+63\\\hline\\&&&&-75&\end{array}$$

Therefore, $$$\frac{2 x^{3} - x^{2} - 12}{x + 3} = \left(2 x^{2} - 7 x + 21\right) + \frac{-75}{x + 3}$$$.