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