Divide using synthetic division $$ ( x^{5}+6x^{4}+13x^{3}+12x^{2}-40x-32 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-4&1&6&13&12&-40&-32\\& & -4& -8& -20& 32& \color{black}{32} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{-8}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}+6x^{4}+13x^{3}+12x^{2}-40x-32 }{ x+4 } = \color{blue}{x^{4}+2x^{3}+5x^{2}-8x-8} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&6&13&12&-40&-32\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-4&\color{orangered}{ 1 }&6&13&12&-40&-32\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&6&13&12&-40&-32\\& & \color{blue}{-4} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}-4&1&\color{orangered}{ 6 }&13&12&-40&-32\\& & \color{orangered}{-4} & & & & \\ \hline &1&\color{orangered}{2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&6&13&12&-40&-32\\& & -4& \color{blue}{-8} & & & \\ \hline &1&\color{blue}{2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}-4&1&6&\color{orangered}{ 13 }&12&-40&-32\\& & -4& \color{orangered}{-8} & & & \\ \hline &1&2&\color{orangered}{5}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&6&13&12&-40&-32\\& & -4& -8& \color{blue}{-20} & & \\ \hline &1&2&\color{blue}{5}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}-4&1&6&13&\color{orangered}{ 12 }&-40&-32\\& & -4& -8& \color{orangered}{-20} & & \\ \hline &1&2&5&\color{orangered}{-8}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&6&13&12&-40&-32\\& & -4& -8& -20& \color{blue}{32} & \\ \hline &1&2&5&\color{blue}{-8}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 32 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}-4&1&6&13&12&\color{orangered}{ -40 }&-32\\& & -4& -8& -20& \color{orangered}{32} & \\ \hline &1&2&5&-8&\color{orangered}{-8}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&6&13&12&-40&-32\\& & -4& -8& -20& 32& \color{blue}{32} \\ \hline &1&2&5&-8&\color{blue}{-8}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 32 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-4&1&6&13&12&-40&\color{orangered}{ -32 }\\& & -4& -8& -20& 32& \color{orangered}{32} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{-8}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+2x^{3}+5x^{2}-8x-8 } $ with a remainder of $ \color{red}{ 0 } $.