Divide using synthetic division $$ ( 2x^{4}-x^{3}+4x-2 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-1&0&4&-2\\& & 6& 15& 45& \color{black}{147} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{15}&\color{blue}{49}&\color{orangered}{145} \end{array} $$The solution is:
$$ \dfrac{ 2x^{4}-x^{3}+4x-2 }{ x-3 } = \color{blue}{2x^{3}+5x^{2}+15x+49} ~+~ \dfrac{ \color{red}{ 145 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-1&0&4&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-1&0&4&-2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-1&0&4&-2\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 6 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -1 }&0&4&-2\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-1&0&4&-2\\& & 6& \color{blue}{15} & & \\ \hline &2&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 15 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}3&2&-1&\color{orangered}{ 0 }&4&-2\\& & 6& \color{orangered}{15} & & \\ \hline &2&5&\color{orangered}{15}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-1&0&4&-2\\& & 6& 15& \color{blue}{45} & \\ \hline &2&5&\color{blue}{15}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 45 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrrr}3&2&-1&0&\color{orangered}{ 4 }&-2\\& & 6& 15& \color{orangered}{45} & \\ \hline &2&5&15&\color{orangered}{49}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 49 } = \color{blue}{ 147 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-1&0&4&-2\\& & 6& 15& 45& \color{blue}{147} \\ \hline &2&5&15&\color{blue}{49}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 147 } = \color{orangered}{ 145 } $
$$ \begin{array}{c|rrrrr}3&2&-1&0&4&\color{orangered}{ -2 }\\& & 6& 15& 45& \color{orangered}{147} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{15}&\color{blue}{49}&\color{orangered}{145} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+5x^{2}+15x+49 } $ with a remainder of $ \color{red}{ 145 } $.