Divide using synthetic division $$ ( x^{3}-3x^{2}+2 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&-3&0&2\\& & -2& 10& \color{black}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{10}&\color{orangered}{-18} \end{array} $$The solution is:
$$ \frac{ x^{3}-3x^{2}+2 }{ x+2 } = \color{blue}{x^{2}-5x+10} \color{red}{~-~} \frac{ \color{red}{ 18 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-3&0&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&-3&0&2\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-3&0&2\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ -3 }&0&2\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-3&0&2\\& & -2& \color{blue}{10} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-2&1&-3&\color{orangered}{ 0 }&2\\& & -2& \color{orangered}{10} & \\ \hline &1&-5&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 10 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-3&0&2\\& & -2& 10& \color{blue}{-20} \\ \hline &1&-5&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}-2&1&-3&0&\color{orangered}{ 2 }\\& & -2& 10& \color{orangered}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{10}&\color{orangered}{-18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x+10 } $ with a remainder of $ \color{red}{ -18 } $.