Divide using synthetic division $$ ( x^{5}-2x^{3}+2x ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&0&-2&0&2&0\\& & -3& 9& -21& 63& \color{black}{-195} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{7}&\color{blue}{-21}&\color{blue}{65}&\color{orangered}{-195} \end{array} $$The solution is:
$$ \frac{ x^{5}-2x^{3}+2x }{ x+3 } = \color{blue}{x^{4}-3x^{3}+7x^{2}-21x+65} \color{red}{~-~} \frac{ \color{red}{ 195 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-2&0&2&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&0&-2&0&2&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-2&0&2&0\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 0 }&-2&0&2&0\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-2&0&2&0\\& & -3& \color{blue}{9} & & & \\ \hline &1&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 9 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&\color{orangered}{ -2 }&0&2&0\\& & -3& \color{orangered}{9} & & & \\ \hline &1&-3&\color{orangered}{7}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-2&0&2&0\\& & -3& 9& \color{blue}{-21} & & \\ \hline &1&-3&\color{blue}{7}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -21 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-2&\color{orangered}{ 0 }&2&0\\& & -3& 9& \color{orangered}{-21} & & \\ \hline &1&-3&7&\color{orangered}{-21}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -21 \right) } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-2&0&2&0\\& & -3& 9& -21& \color{blue}{63} & \\ \hline &1&-3&7&\color{blue}{-21}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 63 } = \color{orangered}{ 65 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-2&0&\color{orangered}{ 2 }&0\\& & -3& 9& -21& \color{orangered}{63} & \\ \hline &1&-3&7&-21&\color{orangered}{65}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 65 } = \color{blue}{ -195 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-2&0&2&0\\& & -3& 9& -21& 63& \color{blue}{-195} \\ \hline &1&-3&7&-21&\color{blue}{65}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -195 \right) } = \color{orangered}{ -195 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-2&0&2&\color{orangered}{ 0 }\\& & -3& 9& -21& 63& \color{orangered}{-195} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{7}&\color{blue}{-21}&\color{blue}{65}&\color{orangered}{-195} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-3x^{3}+7x^{2}-21x+65 } $ with a remainder of $ \color{red}{ -195 } $.