Divide using synthetic division $$ ( x^{5}+17x^{4}+72x^{3}-4 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-8&1&17&72&0&0&-4\\& & -8& -72& 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ x^{5}+17x^{4}+72x^{3}-4 }{ x+8 } = \color{blue}{x^{4}+9x^{3}} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x+8 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 8 = 0 $ ( $ x = \color{blue}{ -8 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&1&17&72&0&0&-4\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-8&\color{orangered}{ 1 }&17&72&0&0&-4\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 1 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&1&17&72&0&0&-4\\& & \color{blue}{-8} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}-8&1&\color{orangered}{ 17 }&72&0&0&-4\\& & \color{orangered}{-8} & & & & \\ \hline &1&\color{orangered}{9}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 9 } = \color{blue}{ -72 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&1&17&72&0&0&-4\\& & -8& \color{blue}{-72} & & & \\ \hline &1&\color{blue}{9}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 72 } + \color{orangered}{ \left( -72 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-8&1&17&\color{orangered}{ 72 }&0&0&-4\\& & -8& \color{orangered}{-72} & & & \\ \hline &1&9&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&1&17&72&0&0&-4\\& & -8& -72& \color{blue}{0} & & \\ \hline &1&9&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-8&1&17&72&\color{orangered}{ 0 }&0&-4\\& & -8& -72& \color{orangered}{0} & & \\ \hline &1&9&0&\color{orangered}{0}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&1&17&72&0&0&-4\\& & -8& -72& 0& \color{blue}{0} & \\ \hline &1&9&0&\color{blue}{0}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-8&1&17&72&0&\color{orangered}{ 0 }&-4\\& & -8& -72& 0& \color{orangered}{0} & \\ \hline &1&9&0&0&\color{orangered}{0}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&1&17&72&0&0&-4\\& & -8& -72& 0& 0& \color{blue}{0} \\ \hline &1&9&0&0&\color{blue}{0}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 0 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}-8&1&17&72&0&0&\color{orangered}{ -4 }\\& & -8& -72& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+9x^{3} } $ with a remainder of $ \color{red}{ -4 } $.