Divide using synthetic division $$ ( x^{5}-7x^{3}+x ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-4&1&0&-7&0&1&0\\& & -4& 16& -36& 144& \color{black}{-580} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{9}&\color{blue}{-36}&\color{blue}{145}&\color{orangered}{-580} \end{array} $$The solution is:
$$ \frac{ x^{5}-7x^{3}+x }{ x+4 } = \color{blue}{x^{4}-4x^{3}+9x^{2}-36x+145} \color{red}{~-~} \frac{ \color{red}{ 580 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&0&-7&0&1&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-4&\color{orangered}{ 1 }&0&-7&0&1&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&0&-7&0&1&0\\& & \color{blue}{-4} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}-4&1&\color{orangered}{ 0 }&-7&0&1&0\\& & \color{orangered}{-4} & & & & \\ \hline &1&\color{orangered}{-4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&0&-7&0&1&0\\& & -4& \color{blue}{16} & & & \\ \hline &1&\color{blue}{-4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 16 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}-4&1&0&\color{orangered}{ -7 }&0&1&0\\& & -4& \color{orangered}{16} & & & \\ \hline &1&-4&\color{orangered}{9}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 9 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&0&-7&0&1&0\\& & -4& 16& \color{blue}{-36} & & \\ \hline &1&-4&\color{blue}{9}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -36 } $
$$ \begin{array}{c|rrrrrr}-4&1&0&-7&\color{orangered}{ 0 }&1&0\\& & -4& 16& \color{orangered}{-36} & & \\ \hline &1&-4&9&\color{orangered}{-36}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -36 \right) } = \color{blue}{ 144 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&0&-7&0&1&0\\& & -4& 16& -36& \color{blue}{144} & \\ \hline &1&-4&9&\color{blue}{-36}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 144 } = \color{orangered}{ 145 } $
$$ \begin{array}{c|rrrrrr}-4&1&0&-7&0&\color{orangered}{ 1 }&0\\& & -4& 16& -36& \color{orangered}{144} & \\ \hline &1&-4&9&-36&\color{orangered}{145}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 145 } = \color{blue}{ -580 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&1&0&-7&0&1&0\\& & -4& 16& -36& 144& \color{blue}{-580} \\ \hline &1&-4&9&-36&\color{blue}{145}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -580 \right) } = \color{orangered}{ -580 } $
$$ \begin{array}{c|rrrrrr}-4&1&0&-7&0&1&\color{orangered}{ 0 }\\& & -4& 16& -36& 144& \color{orangered}{-580} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{9}&\color{blue}{-36}&\color{blue}{145}&\color{orangered}{-580} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-4x^{3}+9x^{2}-36x+145 } $ with a remainder of $ \color{red}{ -580 } $.