Divide using synthetic division $$ ( 6x^{3}+7x^{2}-9x+2 ) \div ( 3x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&6&7&-9&2\\& & 6& 13& \color{black}{4} \\ \hline &\color{blue}{6}&\color{blue}{13}&\color{blue}{4}&\color{orangered}{6} \end{array} $$The solution is:
$$ \dfrac{ 6x^{3}+7x^{2}-9x+2 }{ x-1 } = \color{blue}{6x^{2}+13x+4} ~+~ \dfrac{ \color{red}{ 6 } }{ x-1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&7&-9&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 6 }&7&-9&2\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 6 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&7&-9&2\\& & \color{blue}{6} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 6 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}1&6&\color{orangered}{ 7 }&-9&2\\& & \color{orangered}{6} & & \\ \hline &6&\color{orangered}{13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 13 } = \color{blue}{ 13 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&7&-9&2\\& & 6& \color{blue}{13} & \\ \hline &6&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 13 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}1&6&7&\color{orangered}{ -9 }&2\\& & 6& \color{orangered}{13} & \\ \hline &6&13&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&7&-9&2\\& & 6& 13& \color{blue}{4} \\ \hline &6&13&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 4 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}1&6&7&-9&\color{orangered}{ 2 }\\& & 6& 13& \color{orangered}{4} \\ \hline &\color{blue}{6}&\color{blue}{13}&\color{blue}{4}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+13x+4 } $ with a remainder of $ \color{red}{ 6 } $.