Divide using synthetic division $$ ( 4x^{3}-49x-60 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&0&-49&-60\\& & 12& 36& \color{black}{-39} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{-13}&\color{orangered}{-99} \end{array} $$The solution is:
$$ \dfrac{ 4x^{3}-49x-60 }{ x-3 } = \color{blue}{4x^{2}+12x-13} \color{red}{~-~} \dfrac{ \color{red}{ 99 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-49&-60\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&0&-49&-60\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-49&-60\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ 0 }&-49&-60\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-49&-60\\& & 12& \color{blue}{36} & \\ \hline &4&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -49 } + \color{orangered}{ 36 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}3&4&0&\color{orangered}{ -49 }&-60\\& & 12& \color{orangered}{36} & \\ \hline &4&12&\color{orangered}{-13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-49&-60\\& & 12& 36& \color{blue}{-39} \\ \hline &4&12&\color{blue}{-13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ \left( -39 \right) } = \color{orangered}{ -99 } $
$$ \begin{array}{c|rrrr}3&4&0&-49&\color{orangered}{ -60 }\\& & 12& 36& \color{orangered}{-39} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{-13}&\color{orangered}{-99} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+12x-13 } $ with a remainder of $ \color{red}{ -99 } $.