Check if $ x-8 $ is a factor of $ x^{3}+3x^{2}-10x+24 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&1&3&-10&24\\& & 8& 88& \color{black}{624} \\ \hline &\color{blue}{1}&\color{blue}{11}&\color{blue}{78}&\color{orangered}{648} \end{array} $$Because the remainder $ \left( \color{red}{ 648 } \right) $ is not zero, we conclude that the $ x-8 $ 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 -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&3&-10&24\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 1 }&3&-10&24\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&3&-10&24\\& & \color{blue}{8} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 8 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}8&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{8} & & \\ \hline &1&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 11 } = \color{blue}{ 88 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&3&-10&24\\& & 8& \color{blue}{88} & \\ \hline &1&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 88 } = \color{orangered}{ 78 } $
$$ \begin{array}{c|rrrr}8&1&3&\color{orangered}{ -10 }&24\\& & 8& \color{orangered}{88} & \\ \hline &1&11&\color{orangered}{78}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 78 } = \color{blue}{ 624 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&3&-10&24\\& & 8& 88& \color{blue}{624} \\ \hline &1&11&\color{blue}{78}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ 624 } = \color{orangered}{ 648 } $
$$ \begin{array}{c|rrrr}8&1&3&-10&\color{orangered}{ 24 }\\& & 8& 88& \color{orangered}{624} \\ \hline &\color{blue}{1}&\color{blue}{11}&\color{blue}{78}&\color{orangered}{648} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 648 }\right)$.