Check if $ x-3 $ is a factor of $ x^{3}+3x^{2}-10x+24 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&3&-10&24\\& & 3& 18& \color{black}{24} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{8}&\color{orangered}{48} \end{array} $$Because the remainder $ \left( \color{red}{ 48 } \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}{ 3 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&3&-10&24\\& & 3& \color{blue}{18} & \\ \hline &1&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 18 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}3&1&3&\color{orangered}{ -10 }&24\\& & 3& \color{orangered}{18} & \\ \hline &1&6&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&3&-10&24\\& & 3& 18& \color{blue}{24} \\ \hline &1&6&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ 24 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}3&1&3&-10&\color{orangered}{ 24 }\\& & 3& 18& \color{orangered}{24} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{8}&\color{orangered}{48} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 48 }\right)$.