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