The synthetic division table is:
$$ \begin{array}{c|rrr}2&2&3&-4\\& & 4& \color{black}{14} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{orangered}{10} \end{array} $$Because the remainder $ \left( \color{red}{ 10 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ 2x^{2}+3x-4$.
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.
$$ \begin{array}{c|rrr}\color{blue}{2}&2&3&-4\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}2&\color{orangered}{ 2 }&3&-4\\& & & \\ \hline &\color{orangered}{2}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{2}&2&3&-4\\& & \color{blue}{4} & \\ \hline &\color{blue}{2}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 4 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrr}2&2&\color{orangered}{ 3 }&-4\\& & \color{orangered}{4} & \\ \hline &2&\color{orangered}{7}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrr}\color{blue}{2}&2&3&-4\\& & 4& \color{blue}{14} \\ \hline &2&\color{blue}{7}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 14 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrr}2&2&3&\color{orangered}{ -4 }\\& & 4& \color{orangered}{14} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{orangered}{10} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 10 }\right)$.