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

The synthetic division table is:

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

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

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.

$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&0&-3&2\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&0&-3&2\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

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 } $.

$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&0&-3&2\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $

$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 0 }&-3&2\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{-2}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&0&-3&2\\& & -2& \color{blue}{4} & \\ \hline &1&\color{blue}{-2}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}-2&1&0&\color{orangered}{ -3 }&2\\& & -2& \color{orangered}{4} & \\ \hline &1&-2&\color{orangered}{1}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.

$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&0&-3&2\\& & -2& 4& \color{blue}{-2} \\ \hline &1&-2&\color{blue}{1}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $

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

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

Synthetic Division Calculator