Check if $ x-5 $ is a factor of $ x^{4}+x^{3}+6x^{2}-x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&1&6&-1&5\\& & 5& 30& 180& \color{black}{895} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{36}&\color{blue}{179}&\color{orangered}{900} \end{array} $$Because the remainder $ \left( \color{red}{ 900 } \right) $ is not zero, we conclude that the $ x-5 $ is not a factor of $ x^{4}+x^{3}+6x^{2}-x+5$.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&6&-1&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&1&6&-1&5\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&6&-1&5\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 5 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ 1 }&6&-1&5\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&6&-1&5\\& & 5& \color{blue}{30} & & \\ \hline &1&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 30 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrr}5&1&1&\color{orangered}{ 6 }&-1&5\\& & 5& \color{orangered}{30} & & \\ \hline &1&6&\color{orangered}{36}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 36 } = \color{blue}{ 180 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&6&-1&5\\& & 5& 30& \color{blue}{180} & \\ \hline &1&6&\color{blue}{36}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 180 } = \color{orangered}{ 179 } $
$$ \begin{array}{c|rrrrr}5&1&1&6&\color{orangered}{ -1 }&5\\& & 5& 30& \color{orangered}{180} & \\ \hline &1&6&36&\color{orangered}{179}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 179 } = \color{blue}{ 895 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&6&-1&5\\& & 5& 30& 180& \color{blue}{895} \\ \hline &1&6&36&\color{blue}{179}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 895 } = \color{orangered}{ 900 } $
$$ \begin{array}{c|rrrrr}5&1&1&6&-1&\color{orangered}{ 5 }\\& & 5& 30& 180& \color{orangered}{895} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{36}&\color{blue}{179}&\color{orangered}{900} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 900 }\right)$.