Divide using synthetic division $$ ( 6x^{3}+9x^{2}+9x-4 ) \div ( 6x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&6&9&9&-4\\& & -18& 27& \color{black}{-108} \\ \hline &\color{blue}{6}&\color{blue}{-9}&\color{blue}{36}&\color{orangered}{-112} \end{array} $$The solution is:
$$ \dfrac{ 6x^{3}+9x^{2}+9x-4 }{ x+3 } = \color{blue}{6x^{2}-9x+36} \color{red}{~-~} \dfrac{ \color{red}{ 112 } }{ x+3 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&9&9&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 6 }&9&9&-4\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&9&9&-4\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-3&6&\color{orangered}{ 9 }&9&-4\\& & \color{orangered}{-18} & & \\ \hline &6&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&9&9&-4\\& & -18& \color{blue}{27} & \\ \hline &6&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 27 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrr}-3&6&9&\color{orangered}{ 9 }&-4\\& & -18& \color{orangered}{27} & \\ \hline &6&-9&\color{orangered}{36}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 36 } = \color{blue}{ -108 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&9&9&-4\\& & -18& 27& \color{blue}{-108} \\ \hline &6&-9&\color{blue}{36}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -108 \right) } = \color{orangered}{ -112 } $
$$ \begin{array}{c|rrrr}-3&6&9&9&\color{orangered}{ -4 }\\& & -18& 27& \color{orangered}{-108} \\ \hline &\color{blue}{6}&\color{blue}{-9}&\color{blue}{36}&\color{orangered}{-112} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-9x+36 } $ with a remainder of $ \color{red}{ -112 } $.