Divide using synthetic division $$ ( x^{3}+4x^{2}+x-6 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&4&1&-6\\& & 2& 12& \color{black}{26} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{13}&\color{orangered}{20} \end{array} $$The solution is:
$$ \frac{ x^{3}+4x^{2}+x-6 }{ x-2 } = \color{blue}{x^{2}+6x+13} ~+~ \frac{ \color{red}{ 20 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&4&1&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&4&1&-6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&4&1&-6\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 2 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ 4 }&1&-6\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&4&1&-6\\& & 2& \color{blue}{12} & \\ \hline &1&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 12 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}2&1&4&\color{orangered}{ 1 }&-6\\& & 2& \color{orangered}{12} & \\ \hline &1&6&\color{orangered}{13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 13 } = \color{blue}{ 26 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&4&1&-6\\& & 2& 12& \color{blue}{26} \\ \hline &1&6&\color{blue}{13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 26 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}2&1&4&1&\color{orangered}{ -6 }\\& & 2& 12& \color{orangered}{26} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{13}&\color{orangered}{20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+6x+13 } $ with a remainder of $ \color{red}{ 20 } $.