Divide using synthetic division $$ ( x^{4}+2x^{3}-4x-7 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&2&0&-4&-7\\& & -3& 3& -9& \color{black}{39} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{3}&\color{blue}{-13}&\color{orangered}{32} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}-4x-7 }{ x+3 } = \color{blue}{x^{3}-x^{2}+3x-13} ~+~ \frac{ \color{red}{ 32 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&2&0&-4&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&2&0&-4&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&2&0&-4&-7\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 2 }&0&-4&-7\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&2&0&-4&-7\\& & -3& \color{blue}{3} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-3&1&2&\color{orangered}{ 0 }&-4&-7\\& & -3& \color{orangered}{3} & & \\ \hline &1&-1&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&2&0&-4&-7\\& & -3& 3& \color{blue}{-9} & \\ \hline &1&-1&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}-3&1&2&0&\color{orangered}{ -4 }&-7\\& & -3& 3& \color{orangered}{-9} & \\ \hline &1&-1&3&\color{orangered}{-13}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&2&0&-4&-7\\& & -3& 3& -9& \color{blue}{39} \\ \hline &1&-1&3&\color{blue}{-13}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 39 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}-3&1&2&0&-4&\color{orangered}{ -7 }\\& & -3& 3& -9& \color{orangered}{39} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{3}&\color{blue}{-13}&\color{orangered}{32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-x^{2}+3x-13 } $ with a remainder of $ \color{red}{ 32 } $.