The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-1&-12&3&10\\& & -3& 12& 0& \color{black}{-9} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ x^{4}-x^{3}-12x^{2}+3x+10 }{ x+3 } = \color{blue}{x^{3}-4x^{2}+3} ~+~ \frac{ \color{red}{ 1 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-1&-12&3&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-1&-12&3&10\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-1&-12&3&10\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -1 }&-12&3&10\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-1&-12&3&10\\& & -3& \color{blue}{12} & & \\ \hline &1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&1&-1&\color{orangered}{ -12 }&3&10\\& & -3& \color{orangered}{12} & & \\ \hline &1&-4&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-1&-12&3&10\\& & -3& 12& \color{blue}{0} & \\ \hline &1&-4&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-3&1&-1&-12&\color{orangered}{ 3 }&10\\& & -3& 12& \color{orangered}{0} & \\ \hline &1&-4&0&\color{orangered}{3}& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-1&-12&3&10\\& & -3& 12& 0& \color{blue}{-9} \\ \hline &1&-4&0&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-3&1&-1&-12&3&\color{orangered}{ 10 }\\& & -3& 12& 0& \color{orangered}{-9} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-4x^{2}+3 } $ with a remainder of $ \color{red}{ 1 } $.