Check if $ x+5 $ is a factor of $ x^{3}+x^{2}-19x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&1&-19&8\\& & -5& 20& \color{black}{-5} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{1}&\color{orangered}{3} \end{array} $$Because the remainder $ \left( \color{red}{ 3 } \right) $ is not zero, we conclude that the $ x+5 $ is not a factor of $ x^{3}+x^{2}-19x+8$.
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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&1&-19&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&1&-19&8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&1&-19&8\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ 1 }&-19&8\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&1&-19&8\\& & -5& \color{blue}{20} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 20 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}-5&1&1&\color{orangered}{ -19 }&8\\& & -5& \color{orangered}{20} & \\ \hline &1&-4&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&1&-19&8\\& & -5& 20& \color{blue}{-5} \\ \hline &1&-4&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-5&1&1&-19&\color{orangered}{ 8 }\\& & -5& 20& \color{orangered}{-5} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{1}&\color{orangered}{3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 3 }\right)$.