Divide using synthetic division $$ ( x^{4}-4x^{3}-16x^{2}+22x+18 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&1&-4&-16&22&18\\& & 6& 12& -24& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-4}&\color{blue}{-2}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ x^{4}-4x^{3}-16x^{2}+22x+18 }{ x-6 } = \color{blue}{x^{3}+2x^{2}-4x-2} ~+~ \frac{ \color{red}{ 6 } }{ x-6 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-4&-16&22&18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}6&\color{orangered}{ 1 }&-4&-16&22&18\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-4&-16&22&18\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 6 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}6&1&\color{orangered}{ -4 }&-16&22&18\\& & \color{orangered}{6} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-4&-16&22&18\\& & 6& \color{blue}{12} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 12 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}6&1&-4&\color{orangered}{ -16 }&22&18\\& & 6& \color{orangered}{12} & & \\ \hline &1&2&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-4&-16&22&18\\& & 6& 12& \color{blue}{-24} & \\ \hline &1&2&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}6&1&-4&-16&\color{orangered}{ 22 }&18\\& & 6& 12& \color{orangered}{-24} & \\ \hline &1&2&-4&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-4&-16&22&18\\& & 6& 12& -24& \color{blue}{-12} \\ \hline &1&2&-4&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}6&1&-4&-16&22&\color{orangered}{ 18 }\\& & 6& 12& -24& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-4}&\color{blue}{-2}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}-4x-2 } $ with a remainder of $ \color{red}{ 6 } $.