Find remainder when $ x^{3}-34x-12 $ is divided by $ x-6 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&1&0&-34&-12\\& & 6& 36& \color{black}{12} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The remainder when $ x^{3}-34x-12 $ is divided by $ x-6 $ is $ \, \color{red}{ 0 } $.
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 -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&0&-34&-12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 1 }&0&-34&-12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&0&-34&-12\\& & \color{blue}{6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}6&1&\color{orangered}{ 0 }&-34&-12\\& & \color{orangered}{6} & & \\ \hline &1&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 6 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&0&-34&-12\\& & 6& \color{blue}{36} & \\ \hline &1&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -34 } + \color{orangered}{ 36 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}6&1&0&\color{orangered}{ -34 }&-12\\& & 6& \color{orangered}{36} & \\ \hline &1&6&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&0&-34&-12\\& & 6& 36& \color{blue}{12} \\ \hline &1&6&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}6&1&0&-34&\color{orangered}{ -12 }\\& & 6& 36& \color{orangered}{12} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.