The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&4&-3&-5\\& & -3& -3& \color{black}{18} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-6}&\color{orangered}{13} \end{array} $$The solution is:
$$ \frac{ x^{3}+4x^{2}-3x-5 }{ x+3 } = \color{blue}{x^{2}+x-6} ~+~ \frac{ \color{red}{ 13 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&4&-3&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&4&-3&-5\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&4&-3&-5\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ 4 }&-3&-5\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&4&-3&-5\\& & -3& \color{blue}{-3} & \\ \hline &1&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-3&1&4&\color{orangered}{ -3 }&-5\\& & -3& \color{orangered}{-3} & \\ \hline &1&1&\color{orangered}{-6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&4&-3&-5\\& & -3& -3& \color{blue}{18} \\ \hline &1&1&\color{blue}{-6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 18 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}-3&1&4&-3&\color{orangered}{ -5 }\\& & -3& -3& \color{orangered}{18} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-6}&\color{orangered}{13} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+x-6 } $ with a remainder of $ \color{red}{ 13 } $.