Divide using synthetic division $$ ( 2x^{4}-2 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&0&0&0&-2\\& & 6& 18& 54& \color{black}{162} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{18}&\color{blue}{54}&\color{orangered}{160} \end{array} $$The solution is:
$$ \dfrac{ 2x^{4}-2 }{ x-3 } = \color{blue}{2x^{3}+6x^{2}+18x+54} ~+~ \dfrac{ \color{red}{ 160 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&0&0&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&0&0&0&-2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&0&0&-2\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ 0 }&0&0&-2\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&0&0&-2\\& & 6& \color{blue}{18} & & \\ \hline &2&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 18 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}3&2&0&\color{orangered}{ 0 }&0&-2\\& & 6& \color{orangered}{18} & & \\ \hline &2&6&\color{orangered}{18}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 18 } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&0&0&-2\\& & 6& 18& \color{blue}{54} & \\ \hline &2&6&\color{blue}{18}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 54 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrrr}3&2&0&0&\color{orangered}{ 0 }&-2\\& & 6& 18& \color{orangered}{54} & \\ \hline &2&6&18&\color{orangered}{54}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 54 } = \color{blue}{ 162 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&0&0&-2\\& & 6& 18& 54& \color{blue}{162} \\ \hline &2&6&18&\color{blue}{54}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 162 } = \color{orangered}{ 160 } $
$$ \begin{array}{c|rrrrr}3&2&0&0&0&\color{orangered}{ -2 }\\& & 6& 18& 54& \color{orangered}{162} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{18}&\color{blue}{54}&\color{orangered}{160} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+6x^{2}+18x+54 } $ with a remainder of $ \color{red}{ 160 } $.