Divide using synthetic division $$ ( x^{4}-16x^{3}+18x^{2}-128 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-16&18&0&-128\\& & -2& 36& -108& \color{black}{216} \\ \hline &\color{blue}{1}&\color{blue}{-18}&\color{blue}{54}&\color{blue}{-108}&\color{orangered}{88} \end{array} $$The solution is:
$$ \frac{ x^{4}-16x^{3}+18x^{2}-128 }{ x+2 } = \color{blue}{x^{3}-18x^{2}+54x-108} ~+~ \frac{ \color{red}{ 88 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-16&18&0&-128\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-16&18&0&-128\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-16&18&0&-128\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -16 }&18&0&-128\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-18}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-16&18&0&-128\\& & -2& \color{blue}{36} & & \\ \hline &1&\color{blue}{-18}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ 36 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrrr}-2&1&-16&\color{orangered}{ 18 }&0&-128\\& & -2& \color{orangered}{36} & & \\ \hline &1&-18&\color{orangered}{54}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 54 } = \color{blue}{ -108 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-16&18&0&-128\\& & -2& 36& \color{blue}{-108} & \\ \hline &1&-18&\color{blue}{54}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -108 \right) } = \color{orangered}{ -108 } $
$$ \begin{array}{c|rrrrr}-2&1&-16&18&\color{orangered}{ 0 }&-128\\& & -2& 36& \color{orangered}{-108} & \\ \hline &1&-18&54&\color{orangered}{-108}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -108 \right) } = \color{blue}{ 216 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-16&18&0&-128\\& & -2& 36& -108& \color{blue}{216} \\ \hline &1&-18&54&\color{blue}{-108}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -128 } + \color{orangered}{ 216 } = \color{orangered}{ 88 } $
$$ \begin{array}{c|rrrrr}-2&1&-16&18&0&\color{orangered}{ -128 }\\& & -2& 36& -108& \color{orangered}{216} \\ \hline &\color{blue}{1}&\color{blue}{-18}&\color{blue}{54}&\color{blue}{-108}&\color{orangered}{88} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-18x^{2}+54x-108 } $ with a remainder of $ \color{red}{ 88 } $.