Check if $ x-12 $ is a factor of $ x^{3}+3x^{2}-10x+24 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}12&1&3&-10&24\\& & 12& 180& \color{black}{2040} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{170}&\color{orangered}{2064} \end{array} $$Because the remainder $ \left( \color{red}{ 2064 } \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}{ 12 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}12&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{12} & & \\ \hline &1&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 15 } = \color{blue}{ 180 } $.
$$ \begin{array}{c|rrrr}\color{blue}{12}&1&3&-10&24\\& & 12& \color{blue}{180} & \\ \hline &1&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 180 } = \color{orangered}{ 170 } $
$$ \begin{array}{c|rrrr}12&1&3&\color{orangered}{ -10 }&24\\& & 12& \color{orangered}{180} & \\ \hline &1&15&\color{orangered}{170}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 170 } = \color{blue}{ 2040 } $.
$$ \begin{array}{c|rrrr}\color{blue}{12}&1&3&-10&24\\& & 12& 180& \color{blue}{2040} \\ \hline &1&15&\color{blue}{170}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ 2040 } = \color{orangered}{ 2064 } $
$$ \begin{array}{c|rrrr}12&1&3&-10&\color{orangered}{ 24 }\\& & 12& 180& \color{orangered}{2040} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{170}&\color{orangered}{2064} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 2064 }\right)$.