Divide $$$9 x^{3} + 11 x - 3$$$ by $$$3 x + 2$$$

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

$$$\begin{array}{r|r}\hline\\3 x+2&9 x^{3}+0 x^{2}+11 x-3\end{array}$$$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{9 x^{3}}{3 x} = 3 x^{2}$$$.

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

Multiply it by the divisor: $$$3 x^{2} \left(3 x+2\right) = 9 x^{3}+6 x^{2}$$$.

Subtract the dividend from the obtained result: $$$\left(9 x^{3}+11 x-3\right) - \left(9 x^{3}+6 x^{2}\right) = - 6 x^{2}+11 x-3$$$.

$$\begin{array}{r|rrrr:c}&{\color{Green}3 x^{2}}&&&&\\\hline\\{\color{Magenta}3 x}+2&{\color{Green}9 x^{3}}&+0 x^{2}&+11 x&-3&\frac{{\color{Green}9 x^{3}}}{{\color{Magenta}3 x}} = {\color{Green}3 x^{2}}\\&-\phantom{9 x^{3}}&&&&\\&9 x^{3}&+6 x^{2}&&&{\color{Green}3 x^{2}} \left(3 x+2\right) = 9 x^{3}+6 x^{2}\\\hline\\&&- 6 x^{2}&+11 x&-3&\end{array}$$

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{- 6 x^{2}}{3 x} = - 2 x$$$.

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

Multiply it by the divisor: $$$- 2 x \left(3 x+2\right) = - 6 x^{2}- 4 x$$$.

Subtract the remainder from the obtained result: $$$\left(- 6 x^{2}+11 x-3\right) - \left(- 6 x^{2}- 4 x\right) = 15 x-3$$$.

$$\begin{array}{r|rrrr:c}&3 x^{2}&{\color{DarkMagenta}- 2 x}&&&\\\hline\\{\color{Magenta}3 x}+2&9 x^{3}&+0 x^{2}&+11 x&-3&\\&-\phantom{9 x^{3}}&&&&\\&9 x^{3}&+6 x^{2}&&&\\\hline\\&&{\color{DarkMagenta}- 6 x^{2}}&+11 x&-3&\frac{{\color{DarkMagenta}- 6 x^{2}}}{{\color{Magenta}3 x}} = {\color{DarkMagenta}- 2 x}\\&&-\phantom{- 6 x^{2}}&&&\\&&- 6 x^{2}&- 4 x&&{\color{DarkMagenta}- 2 x} \left(3 x+2\right) = - 6 x^{2}- 4 x\\\hline\\&&&15 x&-3&\end{array}$$

Step 3

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{15 x}{3 x} = 5$$$.

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

Multiply it by the divisor: $$$5 \left(3 x+2\right) = 15 x+10$$$.

Subtract the remainder from the obtained result: $$$\left(15 x-3\right) - \left(15 x+10\right) = -13$$$.

$$\begin{array}{r|rrrr:c}&3 x^{2}&- 2 x&{\color{Fuchsia}+5}&&\\\hline\\{\color{Magenta}3 x}+2&9 x^{3}&+0 x^{2}&+11 x&-3&\\&-\phantom{9 x^{3}}&&&&\\&9 x^{3}&+6 x^{2}&&&\\\hline\\&&- 6 x^{2}&+11 x&-3&\\&&-\phantom{- 6 x^{2}}&&&\\&&- 6 x^{2}&- 4 x&&\\\hline\\&&&{\color{Fuchsia}15 x}&-3&\frac{{\color{Fuchsia}15 x}}{{\color{Magenta}3 x}} = {\color{Fuchsia}5}\\&&&-\phantom{15 x}&&\\&&&15 x&+10&{\color{Fuchsia}5} \left(3 x+2\right) = 15 x+10\\\hline\\&&&&-13&\end{array}$$

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

The resulting table is shown once more:

$$\begin{array}{r|rrrr:c}&{\color{Green}3 x^{2}}&{\color{DarkMagenta}- 2 x}&{\color{Fuchsia}+5}&&\text{Hints}\\\hline\\{\color{Magenta}3 x}+2&{\color{Green}9 x^{3}}&+0 x^{2}&+11 x&-3&\frac{{\color{Green}9 x^{3}}}{{\color{Magenta}3 x}} = {\color{Green}3 x^{2}}\\&-\phantom{9 x^{3}}&&&&\\&9 x^{3}&+6 x^{2}&&&{\color{Green}3 x^{2}} \left(3 x+2\right) = 9 x^{3}+6 x^{2}\\\hline\\&&{\color{DarkMagenta}- 6 x^{2}}&+11 x&-3&\frac{{\color{DarkMagenta}- 6 x^{2}}}{{\color{Magenta}3 x}} = {\color{DarkMagenta}- 2 x}\\&&-\phantom{- 6 x^{2}}&&&\\&&- 6 x^{2}&- 4 x&&{\color{DarkMagenta}- 2 x} \left(3 x+2\right) = - 6 x^{2}- 4 x\\\hline\\&&&{\color{Fuchsia}15 x}&-3&\frac{{\color{Fuchsia}15 x}}{{\color{Magenta}3 x}} = {\color{Fuchsia}5}\\&&&-\phantom{15 x}&&\\&&&15 x&+10&{\color{Fuchsia}5} \left(3 x+2\right) = 15 x+10\\\hline\\&&&&-13&\end{array}$$

Therefore, $$$\frac{9 x^{3} + 11 x - 3}{3 x + 2} = \left(3 x^{2} - 2 x + 5\right) + \frac{-13}{3 x + 2}$$$.