Divide using synthetic division $$ ( x^{4}+8x^{3}+16x^{2}-x-18 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&8&16&-1&-18\\& & -3& -15& -3& \color{black}{12} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{1}&\color{blue}{-4}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \frac{ x^{4}+8x^{3}+16x^{2}-x-18 }{ x+3 } = \color{blue}{x^{3}+5x^{2}+x-4} \color{red}{~-~} \frac{ \color{red}{ 6 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&8&16&-1&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&8&16&-1&-18\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&8&16&-1&-18\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 8 }&16&-1&-18\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&8&16&-1&-18\\& & -3& \color{blue}{-15} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-3&1&8&\color{orangered}{ 16 }&-1&-18\\& & -3& \color{orangered}{-15} & & \\ \hline &1&5&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&8&16&-1&-18\\& & -3& -15& \color{blue}{-3} & \\ \hline &1&5&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-3&1&8&16&\color{orangered}{ -1 }&-18\\& & -3& -15& \color{orangered}{-3} & \\ \hline &1&5&1&\color{orangered}{-4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&8&16&-1&-18\\& & -3& -15& -3& \color{blue}{12} \\ \hline &1&5&1&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 12 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-3&1&8&16&-1&\color{orangered}{ -18 }\\& & -3& -15& -3& \color{orangered}{12} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{1}&\color{blue}{-4}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+x-4 } $ with a remainder of $ \color{red}{ -6 } $.