The synthetic division table is:
$$ \begin{array}{c|rrrr}0&4&6&3&-18\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{3}&\color{orangered}{-18} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+6x^{2}+3x-18 }{ x } = \color{blue}{4x^{2}+6x+3} \color{red}{~-~} \frac{ \color{red}{ 18 } }{ x } $$Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&6&3&-18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 4 }&6&3&-18\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&6&3&-18\\& & \color{blue}{0} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}0&4&\color{orangered}{ 6 }&3&-18\\& & \color{orangered}{0} & & \\ \hline &4&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 6 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&6&3&-18\\& & 0& \color{blue}{0} & \\ \hline &4&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}0&4&6&\color{orangered}{ 3 }&-18\\& & 0& \color{orangered}{0} & \\ \hline &4&6&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 3 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&6&3&-18\\& & 0& 0& \color{blue}{0} \\ \hline &4&6&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 0 } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}0&4&6&3&\color{orangered}{ -18 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{3}&\color{orangered}{-18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+6x+3 } $ with a remainder of $ \color{red}{ -18 } $.