The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&6&-7&-60\\& & 3& 27& \color{black}{60} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{20}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}+6x^{2}-7x-60 }{ x-3 } = \color{blue}{x^{2}+9x+20} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&-7&-60\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&6&-7&-60\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&-7&-60\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 3 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 6 }&-7&-60\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&-7&-60\\& & 3& \color{blue}{27} & \\ \hline &1&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 27 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}3&1&6&\color{orangered}{ -7 }&-60\\& & 3& \color{orangered}{27} & \\ \hline &1&9&\color{orangered}{20}& \end{array} $$Step 6 : 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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&-7&-60\\& & 3& 27& \color{blue}{60} \\ \hline &1&9&\color{blue}{20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ 60 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&1&6&-7&\color{orangered}{ -60 }\\& & 3& 27& \color{orangered}{60} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{20}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+9x+20 } $ with a remainder of $ \color{red}{ 0 } $.