The synthetic division table is:
$$ \begin{array}{c|rrrr}0&2&-1&2&3\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{2}&\color{orangered}{3} \end{array} $$Because the remainder $ \left( \color{red}{ 3 } \right) $ is not zero, we conclude that the $ x $ 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 at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-1&2&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 2 }&-1&2&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-1&2&3\\& & \color{blue}{0} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}0&2&\color{orangered}{ -1 }&2&3\\& & \color{orangered}{0} & & \\ \hline &2&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-1&2&3\\& & 0& \color{blue}{0} & \\ \hline &2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}0&2&-1&\color{orangered}{ 2 }&3\\& & 0& \color{orangered}{0} & \\ \hline &2&-1&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-1&2&3\\& & 0& 0& \color{blue}{0} \\ \hline &2&-1&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}0&2&-1&2&\color{orangered}{ 3 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{2}&\color{orangered}{3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 3 }\right)$.