Divide using synthetic division $$ ( x^{3}+10x^{2}+18 ) \div ( x+6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-6&1&10&0&18\\& & -6& -24& \color{black}{144} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-24}&\color{orangered}{162} \end{array} $$The solution is:
$$ \frac{ x^{3}+10x^{2}+18 }{ x+6 } = \color{blue}{x^{2}+4x-24} ~+~ \frac{ \color{red}{ 162 } }{ x+6 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&10&0&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-6&\color{orangered}{ 1 }&10&0&18\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&10&0&18\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-6&1&\color{orangered}{ 10 }&0&18\\& & \color{orangered}{-6} & & \\ \hline &1&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 4 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&10&0&18\\& & -6& \color{blue}{-24} & \\ \hline &1&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}-6&1&10&\color{orangered}{ 0 }&18\\& & -6& \color{orangered}{-24} & \\ \hline &1&4&\color{orangered}{-24}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ 144 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&10&0&18\\& & -6& -24& \color{blue}{144} \\ \hline &1&4&\color{blue}{-24}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ 144 } = \color{orangered}{ 162 } $
$$ \begin{array}{c|rrrr}-6&1&10&0&\color{orangered}{ 18 }\\& & -6& -24& \color{orangered}{144} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-24}&\color{orangered}{162} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+4x-24 } $ with a remainder of $ \color{red}{ 162 } $.