The synthetic division table is:
$$ \begin{array}{c|rr}-3&20&0\\& & \color{black}{-60} \\ \hline &\color{blue}{20}&\color{orangered}{-60} \end{array} $$The solution is:
$$ \frac{ 20x }{ x+3 } = \color{blue}{20} \color{red}{~-~} \frac{ \color{red}{ 60 } }{ x+3 } $$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.
$$ \begin{array}{c|rr}\color{blue}{-3}&20&0\\& & \\ \hline && \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rr}-3&\color{orangered}{ 20 }&0\\& & \\ \hline &\color{orangered}{20}& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 20 } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rr}\color{blue}{-3}&20&0\\& & \color{blue}{-60} \\ \hline &\color{blue}{20}& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -60 } $
$$ \begin{array}{c|rr}-3&20&\color{orangered}{ 0 }\\& & \color{orangered}{-60} \\ \hline &\color{blue}{20}&\color{orangered}{-60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 20 } $ with a remainder of $ \color{red}{ -60 } $.