Check if $ x+6 $ is a factor of $ x^{3}+3x^{2}-10x+24 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-6&1&3&-10&24\\& & -6& 18& \color{black}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{8}&\color{orangered}{-24} \end{array} $$Because the remainder $ \left( \color{red}{ -24 } \right) $ is not zero, we conclude that the $ x+6 $ is not a factor of $ x^{3}+3x^{2}-10x+24$.
Explanation
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 + 6 = 0 $ ( $ x = \color{blue}{ -6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&3&-10&24\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-6&\color{orangered}{ 1 }&3&-10&24\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 1 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&3&-10&24\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-6&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{-6} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&3&-10&24\\& & -6& \color{blue}{18} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 18 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-6&1&3&\color{orangered}{ -10 }&24\\& & -6& \color{orangered}{18} & \\ \hline &1&-3&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 8 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&3&-10&24\\& & -6& 18& \color{blue}{-48} \\ \hline &1&-3&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}-6&1&3&-10&\color{orangered}{ 24 }\\& & -6& 18& \color{orangered}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{8}&\color{orangered}{-24} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -24 }\right)$.