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