Check if $ x+8 $ is a factor of $ x^{3}+3x^{2}-10x+24 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-8&1&3&-10&24\\& & -8& 40& \color{black}{-240} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{30}&\color{orangered}{-216} \end{array} $$Because the remainder $ \left( \color{red}{ -216 } \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}{ \left( -8 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-8&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{-8} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&1&3&-10&24\\& & -8& \color{blue}{40} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 40 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrr}-8&1&3&\color{orangered}{ -10 }&24\\& & -8& \color{orangered}{40} & \\ \hline &1&-5&\color{orangered}{30}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 30 } = \color{blue}{ -240 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&1&3&-10&24\\& & -8& 40& \color{blue}{-240} \\ \hline &1&-5&\color{blue}{30}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -240 \right) } = \color{orangered}{ -216 } $
$$ \begin{array}{c|rrrr}-8&1&3&-10&\color{orangered}{ 24 }\\& & -8& 40& \color{orangered}{-240} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{30}&\color{orangered}{-216} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -216 }\right)$.