The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&-1&2&0&6\\& & 3& -15& \color{black}{45} \\ \hline &\color{blue}{-1}&\color{blue}{5}&\color{blue}{-15}&\color{orangered}{51} \end{array} $$The solution is:
$$ \frac{ -x^{3}+2x^{2}+6 }{ x+3 } = \color{blue}{-x^{2}+5x-15} ~+~ \frac{ \color{red}{ 51 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&2&0&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ -1 }&2&0&6\\& & & & \\ \hline &\color{orangered}{-1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&2&0&6\\& & \color{blue}{3} & & \\ \hline &\color{blue}{-1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-3&-1&\color{orangered}{ 2 }&0&6\\& & \color{orangered}{3} & & \\ \hline &-1&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&2&0&6\\& & 3& \color{blue}{-15} & \\ \hline &-1&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-3&-1&2&\color{orangered}{ 0 }&6\\& & 3& \color{orangered}{-15} & \\ \hline &-1&5&\color{orangered}{-15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&2&0&6\\& & 3& -15& \color{blue}{45} \\ \hline &-1&5&\color{blue}{-15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 45 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrr}-3&-1&2&0&\color{orangered}{ 6 }\\& & 3& -15& \color{orangered}{45} \\ \hline &\color{blue}{-1}&\color{blue}{5}&\color{blue}{-15}&\color{orangered}{51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{2}+5x-15 } $ with a remainder of $ \color{red}{ 51 } $.