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

The synthetic division table is:

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

Because the remainder $ \left( \color{red}{ -2 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ x^{3}-5x^{2}-x+3$.

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.

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&-5&-1&3\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&-5&-1&3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&-5&-1&3\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 1 } = \color{orangered}{ -4 } $

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

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

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

Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -5 } $

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

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.

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

Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -2 } $

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

Remainder is the last entry in the bottom row $ \left(\color{red}{ -2 }\right)$.

Synthetic Division Calculator