The synthetic division table is:
$$ \begin{array}{c|rrrr}1&6&-8&4&-1\\& & 6& -2& \color{black}{2} \\ \hline &\color{blue}{6}&\color{blue}{-2}&\color{blue}{2}&\color{orangered}{1} \end{array} $$The remainder when $ 6x^{3}-8x^{2}+4x-1 $ is divided by $ x-1 $ is $ \, \color{red}{ 1 } $.
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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&-8&4&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 6 }&-8&4&-1\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&-8&4&-1\\& & \color{blue}{6} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 6 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}1&6&\color{orangered}{ -8 }&4&-1\\& & \color{orangered}{6} & & \\ \hline &6&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&-8&4&-1\\& & 6& \color{blue}{-2} & \\ \hline &6&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}1&6&-8&\color{orangered}{ 4 }&-1\\& & 6& \color{orangered}{-2} & \\ \hline &6&-2&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&6&-8&4&-1\\& & 6& -2& \color{blue}{2} \\ \hline &6&-2&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 2 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}1&6&-8&4&\color{orangered}{ -1 }\\& & 6& -2& \color{orangered}{2} \\ \hline &\color{blue}{6}&\color{blue}{-2}&\color{blue}{2}&\color{orangered}{1} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 1 }\right) $.