Divide using synthetic division $$ ( x^{4}-3x^{3}-3x^{2}-4x+2 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-3&-3&-4&2\\& & -3& 18& -45& \color{black}{147} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{15}&\color{blue}{-49}&\color{orangered}{149} \end{array} $$The solution is:
$$ \frac{ x^{4}-3x^{3}-3x^{2}-4x+2 }{ x+3 } = \color{blue}{x^{3}-6x^{2}+15x-49} ~+~ \frac{ \color{red}{ 149 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-3&-4&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-3&-3&-4&2\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-3&-4&2\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -3 }&-3&-4&2\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-3&-4&2\\& & -3& \color{blue}{18} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 18 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&\color{orangered}{ -3 }&-4&2\\& & -3& \color{orangered}{18} & & \\ \hline &1&-6&\color{orangered}{15}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 15 } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-3&-4&2\\& & -3& 18& \color{blue}{-45} & \\ \hline &1&-6&\color{blue}{15}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ -49 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&-3&\color{orangered}{ -4 }&2\\& & -3& 18& \color{orangered}{-45} & \\ \hline &1&-6&15&\color{orangered}{-49}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -49 \right) } = \color{blue}{ 147 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-3&-4&2\\& & -3& 18& -45& \color{blue}{147} \\ \hline &1&-6&15&\color{blue}{-49}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 147 } = \color{orangered}{ 149 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&-3&-4&\color{orangered}{ 2 }\\& & -3& 18& -45& \color{orangered}{147} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{15}&\color{blue}{-49}&\color{orangered}{149} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+15x-49 } $ with a remainder of $ \color{red}{ 149 } $.