Check if $ x-1 $ is a factor of $ 2x^{4}-x^{3}-8x+4 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}1&2&-1&0&-8&4\\& & 2& 1& 1& \color{black}{-7} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{1}&\color{blue}{-7}&\color{orangered}{-3} \end{array} $$

Because the remainder $ \left( \color{red}{ -3 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ 2x^{4}-x^{3}-8x+4$.

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.

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

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

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

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

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

Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 2 } = \color{orangered}{ 1 } $

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

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

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

Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 1 } = \color{orangered}{ 1 } $

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

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

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

Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 1 } = \color{orangered}{ -7 } $

$$ \begin{array}{c|rrrrr}1&2&-1&0&\color{orangered}{ -8 }&4\\& & 2& 1& \color{orangered}{1} & \\ \hline &2&1&1&\color{orangered}{-7}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -7 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-1&0&-8&4\\& & 2& 1& 1& \color{blue}{-7} \\ \hline &2&1&1&\color{blue}{-7}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -3 } $

$$ \begin{array}{c|rrrrr}1&2&-1&0&-8&\color{orangered}{ 4 }\\& & 2& 1& 1& \color{orangered}{-7} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{1}&\color{blue}{-7}&\color{orangered}{-3} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ -3 }\right)$.

Synthetic Division Calculator