Divide using synthetic division $$ ( x^{4}+9x^{3}+13x^{2}-33x-16 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&9&13&-33&-16\\& & -4& -20& 28& \color{black}{20} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-7}&\color{blue}{-5}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ x^{4}+9x^{3}+13x^{2}-33x-16 }{ x+4 } = \color{blue}{x^{3}+5x^{2}-7x-5} ~+~ \frac{ \color{red}{ 4 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&9&13&-33&-16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&9&13&-33&-16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&9&13&-33&-16\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 9 }&13&-33&-16\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&9&13&-33&-16\\& & -4& \color{blue}{-20} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-4&1&9&\color{orangered}{ 13 }&-33&-16\\& & -4& \color{orangered}{-20} & & \\ \hline &1&5&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&9&13&-33&-16\\& & -4& -20& \color{blue}{28} & \\ \hline &1&5&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -33 } + \color{orangered}{ 28 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-4&1&9&13&\color{orangered}{ -33 }&-16\\& & -4& -20& \color{orangered}{28} & \\ \hline &1&5&-7&\color{orangered}{-5}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&9&13&-33&-16\\& & -4& -20& 28& \color{blue}{20} \\ \hline &1&5&-7&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 20 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-4&1&9&13&-33&\color{orangered}{ -16 }\\& & -4& -20& 28& \color{orangered}{20} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-7}&\color{blue}{-5}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}-7x-5 } $ with a remainder of $ \color{red}{ 4 } $.