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