Divide using synthetic division $$ ( x^{6}-7x^{3}-8 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}2&1&0&0&-7&0&0&-8\\& & 2& 4& 8& 2& 4& \color{black}{8} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{4}&\color{blue}{1}&\color{blue}{2}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{6}-7x^{3}-8 }{ x-2 } = \color{blue}{x^{5}+2x^{4}+4x^{3}+x^{2}+2x+4} $$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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&1&0&0&-7&0&0&-8\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}2&\color{orangered}{ 1 }&0&0&-7&0&0&-8\\& & & & & & & \\ \hline &\color{orangered}{1}&&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&1&0&0&-7&0&0&-8\\& & \color{blue}{2} & & & & & \\ \hline &\color{blue}{1}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrrr}2&1&\color{orangered}{ 0 }&0&-7&0&0&-8\\& & \color{orangered}{2} & & & & & \\ \hline &1&\color{orangered}{2}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&1&0&0&-7&0&0&-8\\& & 2& \color{blue}{4} & & & & \\ \hline &1&\color{blue}{2}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrrr}2&1&0&\color{orangered}{ 0 }&-7&0&0&-8\\& & 2& \color{orangered}{4} & & & & \\ \hline &1&2&\color{orangered}{4}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&1&0&0&-7&0&0&-8\\& & 2& 4& \color{blue}{8} & & & \\ \hline &1&2&\color{blue}{4}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 8 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrrr}2&1&0&0&\color{orangered}{ -7 }&0&0&-8\\& & 2& 4& \color{orangered}{8} & & & \\ \hline &1&2&4&\color{orangered}{1}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&1&0&0&-7&0&0&-8\\& & 2& 4& 8& \color{blue}{2} & & \\ \hline &1&2&4&\color{blue}{1}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrrr}2&1&0&0&-7&\color{orangered}{ 0 }&0&-8\\& & 2& 4& 8& \color{orangered}{2} & & \\ \hline &1&2&4&1&\color{orangered}{2}&& \end{array} $$Step 10 : 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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&1&0&0&-7&0&0&-8\\& & 2& 4& 8& 2& \color{blue}{4} & \\ \hline &1&2&4&1&\color{blue}{2}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrrr}2&1&0&0&-7&0&\color{orangered}{ 0 }&-8\\& & 2& 4& 8& 2& \color{orangered}{4} & \\ \hline &1&2&4&1&2&\color{orangered}{4}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&1&0&0&-7&0&0&-8\\& & 2& 4& 8& 2& 4& \color{blue}{8} \\ \hline &1&2&4&1&2&\color{blue}{4}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 8 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}2&1&0&0&-7&0&0&\color{orangered}{ -8 }\\& & 2& 4& 8& 2& 4& \color{orangered}{8} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{4}&\color{blue}{1}&\color{blue}{2}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{5}+2x^{4}+4x^{3}+x^{2}+2x+4 } $ with a remainder of $ \color{red}{ 0 } $.