Divide using synthetic division $$ ( 6x^{4}+5x^{3}-10x+4 ) \div ( 3x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&6&5&0&-10&4\\& & 12& 34& 68& \color{black}{116} \\ \hline &\color{blue}{6}&\color{blue}{17}&\color{blue}{34}&\color{blue}{58}&\color{orangered}{120} \end{array} $$The solution is:
$$ \dfrac{ 6x^{4}+5x^{3}-10x+4 }{ x-2 } = \color{blue}{6x^{3}+17x^{2}+34x+58} ~+~ \dfrac{ \color{red}{ 120 } }{ x-2 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&5&0&-10&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 6 }&5&0&-10&4\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&5&0&-10&4\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 12 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}2&6&\color{orangered}{ 5 }&0&-10&4\\& & \color{orangered}{12} & & & \\ \hline &6&\color{orangered}{17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 17 } = \color{blue}{ 34 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&5&0&-10&4\\& & 12& \color{blue}{34} & & \\ \hline &6&\color{blue}{17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 34 } = \color{orangered}{ 34 } $
$$ \begin{array}{c|rrrrr}2&6&5&\color{orangered}{ 0 }&-10&4\\& & 12& \color{orangered}{34} & & \\ \hline &6&17&\color{orangered}{34}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 34 } = \color{blue}{ 68 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&5&0&-10&4\\& & 12& 34& \color{blue}{68} & \\ \hline &6&17&\color{blue}{34}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 68 } = \color{orangered}{ 58 } $
$$ \begin{array}{c|rrrrr}2&6&5&0&\color{orangered}{ -10 }&4\\& & 12& 34& \color{orangered}{68} & \\ \hline &6&17&34&\color{orangered}{58}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 58 } = \color{blue}{ 116 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&5&0&-10&4\\& & 12& 34& 68& \color{blue}{116} \\ \hline &6&17&34&\color{blue}{58}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 116 } = \color{orangered}{ 120 } $
$$ \begin{array}{c|rrrrr}2&6&5&0&-10&\color{orangered}{ 4 }\\& & 12& 34& 68& \color{orangered}{116} \\ \hline &\color{blue}{6}&\color{blue}{17}&\color{blue}{34}&\color{blue}{58}&\color{orangered}{120} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+17x^{2}+34x+58 } $ with a remainder of $ \color{red}{ 120 } $.