Check if $ x-2 $ is a factor of $ -4x^{3}+12x^{2}-12x+8 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}2&-4&12&-12&8\\& & -8& 8& \color{black}{-8} \\ \hline &\color{blue}{-4}&\color{blue}{4}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$

Because the remainder equals zero, we conclude that the $ x-2 $ is a factor of the $ -4x^{3}+12x^{2}-12x+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.

$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&12&-12&8\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}2&\color{orangered}{ -4 }&12&-12&8\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&12&-12&8\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 4 } $

$$ \begin{array}{c|rrrr}2&-4&\color{orangered}{ 12 }&-12&8\\& & \color{orangered}{-8} & & \\ \hline &-4&\color{orangered}{4}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&12&-12&8\\& & -8& \color{blue}{8} & \\ \hline &-4&\color{blue}{4}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 8 } = \color{orangered}{ -4 } $

$$ \begin{array}{c|rrrr}2&-4&12&\color{orangered}{ -12 }&8\\& & -8& \color{orangered}{8} & \\ \hline &-4&4&\color{orangered}{-4}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&12&-12&8\\& & -8& 8& \color{blue}{-8} \\ \hline &-4&4&\color{blue}{-4}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrr}2&-4&12&-12&\color{orangered}{ 8 }\\& & -8& 8& \color{orangered}{-8} \\ \hline &\color{blue}{-4}&\color{blue}{4}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$

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

Synthetic Division Calculator