Check if $ x+12 $ is a factor of $ x^{3}+3x^{2}-10x+24 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-12&1&3&-10&24\\& & -12& 108& \color{black}{-1176} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{98}&\color{orangered}{-1152} \end{array} $$Because the remainder $ \left( \color{red}{ -1152 } \right) $ is not zero, we conclude that the $ x+12 $ 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 + 12 = 0 $ ( $ x = \color{blue}{ -12 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&3&-10&24\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-12&\color{orangered}{ 1 }&3&-10&24\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ 1 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&3&-10&24\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-12&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{-12} & & \\ \hline &1&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 108 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&3&-10&24\\& & -12& \color{blue}{108} & \\ \hline &1&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 108 } = \color{orangered}{ 98 } $
$$ \begin{array}{c|rrrr}-12&1&3&\color{orangered}{ -10 }&24\\& & -12& \color{orangered}{108} & \\ \hline &1&-9&\color{orangered}{98}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ 98 } = \color{blue}{ -1176 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&3&-10&24\\& & -12& 108& \color{blue}{-1176} \\ \hline &1&-9&\color{blue}{98}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -1176 \right) } = \color{orangered}{ -1152 } $
$$ \begin{array}{c|rrrr}-12&1&3&-10&\color{orangered}{ 24 }\\& & -12& 108& \color{orangered}{-1176} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{98}&\color{orangered}{-1152} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -1152 }\right)$.