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