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