The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&3&-10&24\\& & -3& 0& \color{black}{30} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-10}&\color{orangered}{54} \end{array} $$Because the remainder $ \left( \color{red}{ 54 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ x^{3}+3x^{2}-10x+24$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&3&-10&24\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&3&-10&24\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&3&-10&24\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&3&-10&24\\& & -3& \color{blue}{0} & \\ \hline &1&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 0 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-3&1&3&\color{orangered}{ -10 }&24\\& & -3& \color{orangered}{0} & \\ \hline &1&0&\color{orangered}{-10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&3&-10&24\\& & -3& 0& \color{blue}{30} \\ \hline &1&0&\color{blue}{-10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ 30 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrr}-3&1&3&-10&\color{orangered}{ 24 }\\& & -3& 0& \color{orangered}{30} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-10}&\color{orangered}{54} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 54 }\right)$.