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