Check if $ x-1 $ is a factor of $ 4x^{4}-x^{3}+5x^{2}-2x-6 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}1&4&-1&5&-2&-6\\& & 4& 3& 8& \color{black}{6} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{8}&\color{blue}{6}&\color{orangered}{0} \end{array} $$

Because the remainder equals zero, we conclude that the $ x-1 $ is a factor of the $ 4x^{4}-x^{3}+5x^{2}-2x-6 $.

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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&5&-2&-6\\& & & & & \\ \hline &&&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&-1&5&-2&-6\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&5&-2&-6\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 4 } = \color{orangered}{ 3 } $

$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ -1 }&5&-2&-6\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{3}&&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&5&-2&-6\\& & 4& \color{blue}{3} & & \\ \hline &4&\color{blue}{3}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 3 } = \color{orangered}{ 8 } $

$$ \begin{array}{c|rrrrr}1&4&-1&\color{orangered}{ 5 }&-2&-6\\& & 4& \color{orangered}{3} & & \\ \hline &4&3&\color{orangered}{8}&& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&5&-2&-6\\& & 4& 3& \color{blue}{8} & \\ \hline &4&3&\color{blue}{8}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 8 } = \color{orangered}{ 6 } $

$$ \begin{array}{c|rrrrr}1&4&-1&5&\color{orangered}{ -2 }&-6\\& & 4& 3& \color{orangered}{8} & \\ \hline &4&3&8&\color{orangered}{6}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 6 } = \color{blue}{ 6 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&5&-2&-6\\& & 4& 3& 8& \color{blue}{6} \\ \hline &4&3&8&\color{blue}{6}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrrr}1&4&-1&5&-2&\color{orangered}{ -6 }\\& & 4& 3& 8& \color{orangered}{6} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{8}&\color{blue}{6}&\color{orangered}{0} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.

Synthetic Division Calculator