Divide using synthetic division $$ ( 3x^{4}-8x^{2}-11x+2 ) \div ( x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&3&0&-8&-11&2\\& & 6& 12& 8& \color{black}{-6} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \dfrac{ 3x^{4}-8x^{2}-11x+2 }{ x-2 } = \color{blue}{3x^{3}+6x^{2}+4x-3} \color{red}{~-~} \dfrac{ \color{red}{ 4 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&0&-8&-11&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 3 }&0&-8&-11&2\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&0&-8&-11&2\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}2&3&\color{orangered}{ 0 }&-8&-11&2\\& & \color{orangered}{6} & & & \\ \hline &3&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&0&-8&-11&2\\& & 6& \color{blue}{12} & & \\ \hline &3&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 12 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&3&0&\color{orangered}{ -8 }&-11&2\\& & 6& \color{orangered}{12} & & \\ \hline &3&6&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&0&-8&-11&2\\& & 6& 12& \color{blue}{8} & \\ \hline &3&6&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 8 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}2&3&0&-8&\color{orangered}{ -11 }&2\\& & 6& 12& \color{orangered}{8} & \\ \hline &3&6&4&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&0&-8&-11&2\\& & 6& 12& 8& \color{blue}{-6} \\ \hline &3&6&4&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}2&3&0&-8&-11&\color{orangered}{ 2 }\\& & 6& 12& 8& \color{orangered}{-6} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+6x^{2}+4x-3 } $ with a remainder of $ \color{red}{ -4 } $.