Divide using synthetic division $$ ( x^{4}+4x^{3}-5x-11 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&4&0&-5&-11\\& & -4& 0& 0& \color{black}{20} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-5}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ x^{4}+4x^{3}-5x-11 }{ x+4 } = \color{blue}{x^{3}-5} ~+~ \frac{ \color{red}{ 9 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&4&0&-5&-11\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&4&0&-5&-11\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&4&0&-5&-11\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 4 }&0&-5&-11\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&4&0&-5&-11\\& & -4& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&1&4&\color{orangered}{ 0 }&-5&-11\\& & -4& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&4&0&-5&-11\\& & -4& 0& \color{blue}{0} & \\ \hline &1&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-4&1&4&0&\color{orangered}{ -5 }&-11\\& & -4& 0& \color{orangered}{0} & \\ \hline &1&0&0&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&4&0&-5&-11\\& & -4& 0& 0& \color{blue}{20} \\ \hline &1&0&0&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 20 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-4&1&4&0&-5&\color{orangered}{ -11 }\\& & -4& 0& 0& \color{orangered}{20} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-5}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-5 } $ with a remainder of $ \color{red}{ 9 } $.