Check if $ x+5 $ is a factor of $ x^{3}+4x^{2}-x-25 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&4&-1&-25\\& & -5& 5& \color{black}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{-45} \end{array} $$Because the remainder $ \left( \color{red}{ -45 } \right) $ is not zero, we conclude that the $ x+5 $ is not a factor of $ x^{3}+4x^{2}-x-25$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&4&-1&-25\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&4&-1&-25\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&4&-1&-25\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ 4 }&-1&-25\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&4&-1&-25\\& & -5& \color{blue}{5} & \\ \hline &1&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 5 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-5&1&4&\color{orangered}{ -1 }&-25\\& & -5& \color{orangered}{5} & \\ \hline &1&-1&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&4&-1&-25\\& & -5& 5& \color{blue}{-20} \\ \hline &1&-1&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -45 } $
$$ \begin{array}{c|rrrr}-5&1&4&-1&\color{orangered}{ -25 }\\& & -5& 5& \color{orangered}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{-45} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -45 }\right)$.