The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}2&3&0&0&0&0&0&5\\& & 6& 12& 24& 48& 96& \color{black}{192} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{12}&\color{blue}{24}&\color{blue}{48}&\color{blue}{96}&\color{orangered}{197} \end{array} $$The remainder when $ 3x^{6}+5 $ is divided by $ x-2 $ is $ \, \color{red}{ 197 } $.
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&3&0&0&0&0&0&5\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}2&\color{orangered}{ 3 }&0&0&0&0&0&5\\& & & & & & & \\ \hline &\color{orangered}{3}&&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&3&0&0&0&0&0&5\\& & \color{blue}{6} & & & & & \\ \hline &\color{blue}{3}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrrr}2&3&\color{orangered}{ 0 }&0&0&0&0&5\\& & \color{orangered}{6} & & & & & \\ \hline &3&\color{orangered}{6}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&3&0&0&0&0&0&5\\& & 6& \color{blue}{12} & & & & \\ \hline &3&\color{blue}{6}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrrr}2&3&0&\color{orangered}{ 0 }&0&0&0&5\\& & 6& \color{orangered}{12} & & & & \\ \hline &3&6&\color{orangered}{12}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&3&0&0&0&0&0&5\\& & 6& 12& \color{blue}{24} & & & \\ \hline &3&6&\color{blue}{12}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrrrr}2&3&0&0&\color{orangered}{ 0 }&0&0&5\\& & 6& 12& \color{orangered}{24} & & & \\ \hline &3&6&12&\color{orangered}{24}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 24 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&3&0&0&0&0&0&5\\& & 6& 12& 24& \color{blue}{48} & & \\ \hline &3&6&12&\color{blue}{24}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 48 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrrrr}2&3&0&0&0&\color{orangered}{ 0 }&0&5\\& & 6& 12& 24& \color{orangered}{48} & & \\ \hline &3&6&12&24&\color{orangered}{48}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 48 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&3&0&0&0&0&0&5\\& & 6& 12& 24& 48& \color{blue}{96} & \\ \hline &3&6&12&24&\color{blue}{48}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 96 } = \color{orangered}{ 96 } $
$$ \begin{array}{c|rrrrrrr}2&3&0&0&0&0&\color{orangered}{ 0 }&5\\& & 6& 12& 24& 48& \color{orangered}{96} & \\ \hline &3&6&12&24&48&\color{orangered}{96}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 96 } = \color{blue}{ 192 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&3&0&0&0&0&0&5\\& & 6& 12& 24& 48& 96& \color{blue}{192} \\ \hline &3&6&12&24&48&\color{blue}{96}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 192 } = \color{orangered}{ 197 } $
$$ \begin{array}{c|rrrrrrr}2&3&0&0&0&0&0&\color{orangered}{ 5 }\\& & 6& 12& 24& 48& 96& \color{orangered}{192} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{12}&\color{blue}{24}&\color{blue}{48}&\color{blue}{96}&\color{orangered}{197} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 197 }\right) $.