Check if $ x-2 $ is a factor of $ x^{3}-6x^{2}+11x-8 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}2&1&-6&11&-8\\& & 2& -8& \color{black}{6} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{3}&\color{orangered}{-2} \end{array} $$

Because the remainder $ \left( \color{red}{ -2 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ x^{3}-6x^{2}+11x-8$.

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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-6&11&-8\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&-6&11&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-6&11&-8\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 2 } = \color{orangered}{ -4 } $

$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ -6 }&11&-8\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-6&11&-8\\& & 2& \color{blue}{-8} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 3 } $

$$ \begin{array}{c|rrrr}2&1&-6&\color{orangered}{ 11 }&-8\\& & 2& \color{orangered}{-8} & \\ \hline &1&-4&\color{orangered}{3}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-6&11&-8\\& & 2& -8& \color{blue}{6} \\ \hline &1&-4&\color{blue}{3}& \end{array} $$

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

$$ \begin{array}{c|rrrr}2&1&-6&11&\color{orangered}{ -8 }\\& & 2& -8& \color{orangered}{6} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{3}&\color{orangered}{-2} \end{array} $$

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

Synthetic Division Calculator