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

The synthetic division table is:

$$ \begin{array}{c|rrrr}2&2&-1&2&3\\& & 4& 6& \color{black}{16} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{19} \end{array} $$

Because the remainder $ \left( \color{red}{ 19 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ 2x^{3}-x^{2}+2x+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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.

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

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

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

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

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

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

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

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.

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

Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 6 } = \color{orangered}{ 8 } $

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

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-1&2&3\\& & 4& 6& \color{blue}{16} \\ \hline &2&3&\color{blue}{8}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 16 } = \color{orangered}{ 19 } $

$$ \begin{array}{c|rrrr}2&2&-1&2&\color{orangered}{ 3 }\\& & 4& 6& \color{orangered}{16} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{19} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 19 }\right)$.

Synthetic Division Calculator