Divide using synthetic division $$ ( x^{4}+4x^{3}+20x^{2}+x-39 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&1&4&20&1&-39\\& & -1& -3& -17& \color{black}{16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{17}&\color{blue}{-16}&\color{orangered}{-23} \end{array} $$The solution is:
$$ \frac{ x^{4}+4x^{3}+20x^{2}+x-39 }{ x+1 } = \color{blue}{x^{3}+3x^{2}+17x-16} \color{red}{~-~} \frac{ \color{red}{ 23 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&4&20&1&-39\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 1 }&4&20&1&-39\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&4&20&1&-39\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-1&1&\color{orangered}{ 4 }&20&1&-39\\& & \color{orangered}{-1} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&4&20&1&-39\\& & -1& \color{blue}{-3} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}-1&1&4&\color{orangered}{ 20 }&1&-39\\& & -1& \color{orangered}{-3} & & \\ \hline &1&3&\color{orangered}{17}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 17 } = \color{blue}{ -17 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&4&20&1&-39\\& & -1& -3& \color{blue}{-17} & \\ \hline &1&3&\color{blue}{17}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -17 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-1&1&4&20&\color{orangered}{ 1 }&-39\\& & -1& -3& \color{orangered}{-17} & \\ \hline &1&3&17&\color{orangered}{-16}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&4&20&1&-39\\& & -1& -3& -17& \color{blue}{16} \\ \hline &1&3&17&\color{blue}{-16}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -39 } + \color{orangered}{ 16 } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrrrr}-1&1&4&20&1&\color{orangered}{ -39 }\\& & -1& -3& -17& \color{orangered}{16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{17}&\color{blue}{-16}&\color{orangered}{-23} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}+17x-16 } $ with a remainder of $ \color{red}{ -23 } $.