Divide using synthetic division $$ ( x^{4}-3x^{3}+2x^{2}-6x ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-3&2&-6&0\\& & -3& 18& -60& \color{black}{198} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{20}&\color{blue}{-66}&\color{orangered}{198} \end{array} $$The solution is:
$$ \frac{ x^{4}-3x^{3}+2x^{2}-6x }{ x+3 } = \color{blue}{x^{3}-6x^{2}+20x-66} ~+~ \frac{ \color{red}{ 198 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&2&-6&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-3&2&-6&0\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&2&-6&0\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -3 }&2&-6&0\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&2&-6&0\\& & -3& \color{blue}{18} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 18 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&\color{orangered}{ 2 }&-6&0\\& & -3& \color{orangered}{18} & & \\ \hline &1&-6&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 20 } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&2&-6&0\\& & -3& 18& \color{blue}{-60} & \\ \hline &1&-6&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -66 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&2&\color{orangered}{ -6 }&0\\& & -3& 18& \color{orangered}{-60} & \\ \hline &1&-6&20&\color{orangered}{-66}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -66 \right) } = \color{blue}{ 198 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&2&-6&0\\& & -3& 18& -60& \color{blue}{198} \\ \hline &1&-6&20&\color{blue}{-66}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 198 } = \color{orangered}{ 198 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&2&-6&\color{orangered}{ 0 }\\& & -3& 18& -60& \color{orangered}{198} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{20}&\color{blue}{-66}&\color{orangered}{198} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+20x-66 } $ with a remainder of $ \color{red}{ 198 } $.