Divide using synthetic division $$ ( 3x^{4}+2x^{3}+4x ) \div ( x-5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&3&2&0&4&0\\& & 15& 85& 425& \color{black}{2145} \\ \hline &\color{blue}{3}&\color{blue}{17}&\color{blue}{85}&\color{blue}{429}&\color{orangered}{2145} \end{array} $$The solution is:
$$ \dfrac{ 3x^{4}+2x^{3}+4x }{ x-5 } = \color{blue}{3x^{3}+17x^{2}+85x+429} ~+~ \dfrac{ \color{red}{ 2145 } }{ x-5 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&2&0&4&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 3 }&2&0&4&0\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&2&0&4&0\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 15 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}5&3&\color{orangered}{ 2 }&0&4&0\\& & \color{orangered}{15} & & & \\ \hline &3&\color{orangered}{17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 17 } = \color{blue}{ 85 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&2&0&4&0\\& & 15& \color{blue}{85} & & \\ \hline &3&\color{blue}{17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 85 } = \color{orangered}{ 85 } $
$$ \begin{array}{c|rrrrr}5&3&2&\color{orangered}{ 0 }&4&0\\& & 15& \color{orangered}{85} & & \\ \hline &3&17&\color{orangered}{85}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 85 } = \color{blue}{ 425 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&2&0&4&0\\& & 15& 85& \color{blue}{425} & \\ \hline &3&17&\color{blue}{85}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 425 } = \color{orangered}{ 429 } $
$$ \begin{array}{c|rrrrr}5&3&2&0&\color{orangered}{ 4 }&0\\& & 15& 85& \color{orangered}{425} & \\ \hline &3&17&85&\color{orangered}{429}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 429 } = \color{blue}{ 2145 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&2&0&4&0\\& & 15& 85& 425& \color{blue}{2145} \\ \hline &3&17&85&\color{blue}{429}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2145 } = \color{orangered}{ 2145 } $
$$ \begin{array}{c|rrrrr}5&3&2&0&4&\color{orangered}{ 0 }\\& & 15& 85& 425& \color{orangered}{2145} \\ \hline &\color{blue}{3}&\color{blue}{17}&\color{blue}{85}&\color{blue}{429}&\color{orangered}{2145} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+17x^{2}+85x+429 } $ with a remainder of $ \color{red}{ 2145 } $.