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