Divide using synthetic division $$ ( 8x^{5}+39x^{4}+29x^{3}+12x^{2}+24x-33 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-4&8&39&29&12&24&-33\\& & -32& -28& -4& -32& \color{black}{32} \\ \hline &\color{blue}{8}&\color{blue}{7}&\color{blue}{1}&\color{blue}{8}&\color{blue}{-8}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ 8x^{5}+39x^{4}+29x^{3}+12x^{2}+24x-33 }{ x+4 } = \color{blue}{8x^{4}+7x^{3}+x^{2}+8x-8} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&8&39&29&12&24&-33\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-4&\color{orangered}{ 8 }&39&29&12&24&-33\\& & & & & & \\ \hline &\color{orangered}{8}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 8 } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&8&39&29&12&24&-33\\& & \color{blue}{-32} & & & & \\ \hline &\color{blue}{8}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 39 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-4&8&\color{orangered}{ 39 }&29&12&24&-33\\& & \color{orangered}{-32} & & & & \\ \hline &8&\color{orangered}{7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 7 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&8&39&29&12&24&-33\\& & -32& \color{blue}{-28} & & & \\ \hline &8&\color{blue}{7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}-4&8&39&\color{orangered}{ 29 }&12&24&-33\\& & -32& \color{orangered}{-28} & & & \\ \hline &8&7&\color{orangered}{1}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&8&39&29&12&24&-33\\& & -32& -28& \color{blue}{-4} & & \\ \hline &8&7&\color{blue}{1}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}-4&8&39&29&\color{orangered}{ 12 }&24&-33\\& & -32& -28& \color{orangered}{-4} & & \\ \hline &8&7&1&\color{orangered}{8}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 8 } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&8&39&29&12&24&-33\\& & -32& -28& -4& \color{blue}{-32} & \\ \hline &8&7&1&\color{blue}{8}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}-4&8&39&29&12&\color{orangered}{ 24 }&-33\\& & -32& -28& -4& \color{orangered}{-32} & \\ \hline &8&7&1&8&\color{orangered}{-8}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&8&39&29&12&24&-33\\& & -32& -28& -4& -32& \color{blue}{32} \\ \hline &8&7&1&8&\color{blue}{-8}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -33 } + \color{orangered}{ 32 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-4&8&39&29&12&24&\color{orangered}{ -33 }\\& & -32& -28& -4& -32& \color{orangered}{32} \\ \hline &\color{blue}{8}&\color{blue}{7}&\color{blue}{1}&\color{blue}{8}&\color{blue}{-8}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{4}+7x^{3}+x^{2}+8x-8 } $ with a remainder of $ \color{red}{ -1 } $.