The synthetic division table is:
$$ \begin{array}{c|rr}1&2&6\\& & \color{black}{2} \\ \hline &\color{blue}{2}&\color{orangered}{8} \end{array} $$Because the remainder $ \left( \color{red}{ 8 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ 2x+6$.
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.
$$ \begin{array}{c|rr}\color{blue}{1}&2&6\\& & \\ \hline && \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rr}1&\color{orangered}{ 2 }&6\\& & \\ \hline &\color{orangered}{2}& \end{array} $$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 } $.
$$ \begin{array}{c|rr}\color{blue}{1}&2&6\\& & \color{blue}{2} \\ \hline &\color{blue}{2}& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 2 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rr}1&2&\color{orangered}{ 6 }\\& & \color{orangered}{2} \\ \hline &\color{blue}{2}&\color{orangered}{8} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 8 }\right)$.