The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&4&-3&-6\\& & -6& 6& \color{black}{-9} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{-15} \end{array} $$Because the remainder $ \left( \color{red}{ -15 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ 2x^{3}+4x^{2}-3x-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 + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&4&-3&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&4&-3&-6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&4&-3&-6\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ 4 }&-3&-6\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&4&-3&-6\\& & -6& \color{blue}{6} & \\ \hline &2&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 6 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-3&2&4&\color{orangered}{ -3 }&-6\\& & -6& \color{orangered}{6} & \\ \hline &2&-2&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&4&-3&-6\\& & -6& 6& \color{blue}{-9} \\ \hline &2&-2&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-3&2&4&-3&\color{orangered}{ -6 }\\& & -6& 6& \color{orangered}{-9} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{-15} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -15 }\right)$.