Check if $ x-4 $ is a factor of $ 5x^{3}+60x^{2}-5x-1 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&5&60&-5&-1\\& & 20& 320& \color{black}{1260} \\ \hline &\color{blue}{5}&\color{blue}{80}&\color{blue}{315}&\color{orangered}{1259} \end{array} $$Because the remainder $ \left( \color{red}{ 1259 } \right) $ is not zero, we conclude that the $ x-4 $ is not a factor of $ 5x^{3}+60x^{2}-5x-1$.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&60&-5&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 5 }&60&-5&-1\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&60&-5&-1\\& & \color{blue}{20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 60 } + \color{orangered}{ 20 } = \color{orangered}{ 80 } $
$$ \begin{array}{c|rrrr}4&5&\color{orangered}{ 60 }&-5&-1\\& & \color{orangered}{20} & & \\ \hline &5&\color{orangered}{80}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 80 } = \color{blue}{ 320 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&60&-5&-1\\& & 20& \color{blue}{320} & \\ \hline &5&\color{blue}{80}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 320 } = \color{orangered}{ 315 } $
$$ \begin{array}{c|rrrr}4&5&60&\color{orangered}{ -5 }&-1\\& & 20& \color{orangered}{320} & \\ \hline &5&80&\color{orangered}{315}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 315 } = \color{blue}{ 1260 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&60&-5&-1\\& & 20& 320& \color{blue}{1260} \\ \hline &5&80&\color{blue}{315}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 1260 } = \color{orangered}{ 1259 } $
$$ \begin{array}{c|rrrr}4&5&60&-5&\color{orangered}{ -1 }\\& & 20& 320& \color{orangered}{1260} \\ \hline &\color{blue}{5}&\color{blue}{80}&\color{blue}{315}&\color{orangered}{1259} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 1259 }\right)$.