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