The synthetic division table is:
$$ \begin{array}{c|rrr}3&4&2&0\\& & 12& \color{black}{42} \\ \hline &\color{blue}{4}&\color{blue}{14}&\color{orangered}{42} \end{array} $$Because the remainder $ \left( \color{red}{ 42 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ 4x^{2}+2x$.
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&2&0\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 4 }&2&0\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&2&0\\& & \color{blue}{12} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 12 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrr}3&4&\color{orangered}{ 2 }&0\\& & \color{orangered}{12} & \\ \hline &4&\color{orangered}{14}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 14 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&2&0\\& & 12& \color{blue}{42} \\ \hline &4&\color{blue}{14}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 42 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrr}3&4&2&\color{orangered}{ 0 }\\& & 12& \color{orangered}{42} \\ \hline &\color{blue}{4}&\color{blue}{14}&\color{orangered}{42} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 42 }\right)$.