Divide $$$4 x^{4} + 3 x^{3} - 11 x^{2} - x + 2$$$ by $$$x + 2$$$

Write the problem in the special format:

$$$\begin{array}{r|r}\hline\\x+2&4 x^{4}+3 x^{3}- 11 x^{2}- x+2\end{array}$$$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{4 x^{4}}{x} = 4 x^{3}$$$.

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

Multiply it by the divisor: $$$4 x^{3} \left(x+2\right) = 4 x^{4}+8 x^{3}$$$.

Subtract the dividend from the obtained result: $$$\left(4 x^{4}+3 x^{3}- 11 x^{2}- x+2\right) - \left(4 x^{4}+8 x^{3}\right) = - 5 x^{3}- 11 x^{2}- x+2$$$.

$$\begin{array}{r|rrrrr:c}&{\color{Chocolate}4 x^{3}}&&&&&\\\hline\\{\color{Magenta}x}+2&{\color{Chocolate}4 x^{4}}&+3 x^{3}&- 11 x^{2}&- x&+2&\frac{{\color{Chocolate}4 x^{4}}}{{\color{Magenta}x}} = {\color{Chocolate}4 x^{3}}\\&-\phantom{4 x^{4}}&&&&&\\&4 x^{4}&+8 x^{3}&&&&{\color{Chocolate}4 x^{3}} \left(x+2\right) = 4 x^{4}+8 x^{3}\\\hline\\&&- 5 x^{3}&- 11 x^{2}&- x&+2&\end{array}$$

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{- 5 x^{3}}{x} = - 5 x^{2}$$$.

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

Multiply it by the divisor: $$$- 5 x^{2} \left(x+2\right) = - 5 x^{3}- 10 x^{2}$$$.

Subtract the remainder from the obtained result: $$$\left(- 5 x^{3}- 11 x^{2}- x+2\right) - \left(- 5 x^{3}- 10 x^{2}\right) = - x^{2}- x+2$$$.

$$\begin{array}{r|rrrrr:c}&4 x^{3}&{\color{DarkBlue}- 5 x^{2}}&&&&\\\hline\\{\color{Magenta}x}+2&4 x^{4}&+3 x^{3}&- 11 x^{2}&- x&+2&\\&-\phantom{4 x^{4}}&&&&&\\&4 x^{4}&+8 x^{3}&&&&\\\hline\\&&{\color{DarkBlue}- 5 x^{3}}&- 11 x^{2}&- x&+2&\frac{{\color{DarkBlue}- 5 x^{3}}}{{\color{Magenta}x}} = {\color{DarkBlue}- 5 x^{2}}\\&&-\phantom{- 5 x^{3}}&&&&\\&&- 5 x^{3}&- 10 x^{2}&&&{\color{DarkBlue}- 5 x^{2}} \left(x+2\right) = - 5 x^{3}- 10 x^{2}\\\hline\\&&&- x^{2}&- x&+2&\end{array}$$

Step 3

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

$$\begin{array}{r|rrrrr:c}&4 x^{3}&- 5 x^{2}&{\color{SaddleBrown}- x}&&&\\\hline\\{\color{Magenta}x}+2&4 x^{4}&+3 x^{3}&- 11 x^{2}&- x&+2&\\&-\phantom{4 x^{4}}&&&&&\\&4 x^{4}&+8 x^{3}&&&&\\\hline\\&&- 5 x^{3}&- 11 x^{2}&- x&+2&\\&&-\phantom{- 5 x^{3}}&&&&\\&&- 5 x^{3}&- 10 x^{2}&&&\\\hline\\&&&{\color{SaddleBrown}- x^{2}}&- x&+2&\frac{{\color{SaddleBrown}- x^{2}}}{{\color{Magenta}x}} = {\color{SaddleBrown}- x}\\&&&-\phantom{- x^{2}}&&&\\&&&- x^{2}&- 2 x&&{\color{SaddleBrown}- x} \left(x+2\right) = - x^{2}- 2 x\\\hline\\&&&&x&+2&\end{array}$$

Step 4

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{x}{x} = 1$$$.

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

Multiply it by the divisor: $$$1 \left(x+2\right) = x+2$$$.

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

$$\begin{array}{r|rrrrr:c}&4 x^{3}&- 5 x^{2}&- x&{\color{Chartreuse}+1}&&\\\hline\\{\color{Magenta}x}+2&4 x^{4}&+3 x^{3}&- 11 x^{2}&- x&+2&\\&-\phantom{4 x^{4}}&&&&&\\&4 x^{4}&+8 x^{3}&&&&\\\hline\\&&- 5 x^{3}&- 11 x^{2}&- x&+2&\\&&-\phantom{- 5 x^{3}}&&&&\\&&- 5 x^{3}&- 10 x^{2}&&&\\\hline\\&&&- x^{2}&- x&+2&\\&&&-\phantom{- x^{2}}&&&\\&&&- x^{2}&- 2 x&&\\\hline\\&&&&{\color{Chartreuse}x}&+2&\frac{{\color{Chartreuse}x}}{{\color{Magenta}x}} = {\color{Chartreuse}1}\\&&&&-\phantom{x}&&\\&&&&x&+2&{\color{Chartreuse}1} \left(x+2\right) = x+2\\\hline\\&&&&&0&\end{array}$$

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

The resulting table is shown once more:

$$\begin{array}{r|rrrrr:c}&{\color{Chocolate}4 x^{3}}&{\color{DarkBlue}- 5 x^{2}}&{\color{SaddleBrown}- x}&{\color{Chartreuse}+1}&&\text{Hints}\\\hline\\{\color{Magenta}x}+2&{\color{Chocolate}4 x^{4}}&+3 x^{3}&- 11 x^{2}&- x&+2&\frac{{\color{Chocolate}4 x^{4}}}{{\color{Magenta}x}} = {\color{Chocolate}4 x^{3}}\\&-\phantom{4 x^{4}}&&&&&\\&4 x^{4}&+8 x^{3}&&&&{\color{Chocolate}4 x^{3}} \left(x+2\right) = 4 x^{4}+8 x^{3}\\\hline\\&&{\color{DarkBlue}- 5 x^{3}}&- 11 x^{2}&- x&+2&\frac{{\color{DarkBlue}- 5 x^{3}}}{{\color{Magenta}x}} = {\color{DarkBlue}- 5 x^{2}}\\&&-\phantom{- 5 x^{3}}&&&&\\&&- 5 x^{3}&- 10 x^{2}&&&{\color{DarkBlue}- 5 x^{2}} \left(x+2\right) = - 5 x^{3}- 10 x^{2}\\\hline\\&&&{\color{SaddleBrown}- x^{2}}&- x&+2&\frac{{\color{SaddleBrown}- x^{2}}}{{\color{Magenta}x}} = {\color{SaddleBrown}- x}\\&&&-\phantom{- x^{2}}&&&\\&&&- x^{2}&- 2 x&&{\color{SaddleBrown}- x} \left(x+2\right) = - x^{2}- 2 x\\\hline\\&&&&{\color{Chartreuse}x}&+2&\frac{{\color{Chartreuse}x}}{{\color{Magenta}x}} = {\color{Chartreuse}1}\\&&&&-\phantom{x}&&\\&&&&x&+2&{\color{Chartreuse}1} \left(x+2\right) = x+2\\\hline\\&&&&&0&\end{array}$$

Therefore, $$$\frac{4 x^{4} + 3 x^{3} - 11 x^{2} - x + 2}{x + 2} = \left(4 x^{3} - 5 x^{2} - x + 1\right) + \frac{0}{x + 2} = 4 x^{3} - 5 x^{2} - x + 1.$$$