Check if $ x-1 $ is a factor of $ x^{3}-6x+5 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&1&0&-6&5\\& & 1& 1& \color{black}{-5} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-1 $ is a factor of the $ x^{3}-6x+5 $.
First we need to create a synthetic division table.
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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&0&-6&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&0&-6&5\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&0&-6&5\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 1 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 0 }&-6&5\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&0&-6&5\\& & 1& \color{blue}{1} & \\ \hline &1&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 1 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}1&1&0&\color{orangered}{ -6 }&5\\& & 1& \color{orangered}{1} & \\ \hline &1&1&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&0&-6&5\\& & 1& 1& \color{blue}{-5} \\ \hline &1&1&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&1&0&-6&\color{orangered}{ 5 }\\& & 1& 1& \color{orangered}{-5} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.