The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&3&7&-6&-10\\& & -3& -4& \color{black}{10} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+7x^{2}-6x-10 }{ x+1 } = \color{blue}{3x^{2}+4x-10} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&7&-6&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 3 }&7&-6&-10\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&7&-6&-10\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-1&3&\color{orangered}{ 7 }&-6&-10\\& & \color{orangered}{-3} & & \\ \hline &3&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&7&-6&-10\\& & -3& \color{blue}{-4} & \\ \hline &3&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-1&3&7&\color{orangered}{ -6 }&-10\\& & -3& \color{orangered}{-4} & \\ \hline &3&4&\color{orangered}{-10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&7&-6&-10\\& & -3& -4& \color{blue}{10} \\ \hline &3&4&\color{blue}{-10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-1&3&7&-6&\color{orangered}{ -10 }\\& & -3& -4& \color{orangered}{10} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+4x-10 } $ with a remainder of $ \color{red}{ 0 } $.