Divide using synthetic division $$ ( 7x^{5}+31 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-1&7&0&0&0&0&31\\& & -7& 7& -7& 7& \color{black}{-7} \\ \hline &\color{blue}{7}&\color{blue}{-7}&\color{blue}{7}&\color{blue}{-7}&\color{blue}{7}&\color{orangered}{24} \end{array} $$The solution is:
$$ \frac{ 7x^{5}+31 }{ x+1 } = \color{blue}{7x^{4}-7x^{3}+7x^{2}-7x+7} ~+~ \frac{ \color{red}{ 24 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&7&0&0&0&0&31\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-1&\color{orangered}{ 7 }&0&0&0&0&31\\& & & & & & \\ \hline &\color{orangered}{7}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 7 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&7&0&0&0&0&31\\& & \color{blue}{-7} & & & & \\ \hline &\color{blue}{7}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-1&7&\color{orangered}{ 0 }&0&0&0&31\\& & \color{orangered}{-7} & & & & \\ \hline &7&\color{orangered}{-7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&7&0&0&0&0&31\\& & -7& \color{blue}{7} & & & \\ \hline &7&\color{blue}{-7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 7 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-1&7&0&\color{orangered}{ 0 }&0&0&31\\& & -7& \color{orangered}{7} & & & \\ \hline &7&-7&\color{orangered}{7}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 7 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&7&0&0&0&0&31\\& & -7& 7& \color{blue}{-7} & & \\ \hline &7&-7&\color{blue}{7}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-1&7&0&0&\color{orangered}{ 0 }&0&31\\& & -7& 7& \color{orangered}{-7} & & \\ \hline &7&-7&7&\color{orangered}{-7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&7&0&0&0&0&31\\& & -7& 7& -7& \color{blue}{7} & \\ \hline &7&-7&7&\color{blue}{-7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 7 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-1&7&0&0&0&\color{orangered}{ 0 }&31\\& & -7& 7& -7& \color{orangered}{7} & \\ \hline &7&-7&7&-7&\color{orangered}{7}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 7 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&7&0&0&0&0&31\\& & -7& 7& -7& 7& \color{blue}{-7} \\ \hline &7&-7&7&-7&\color{blue}{7}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 31 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrrr}-1&7&0&0&0&0&\color{orangered}{ 31 }\\& & -7& 7& -7& 7& \color{orangered}{-7} \\ \hline &\color{blue}{7}&\color{blue}{-7}&\color{blue}{7}&\color{blue}{-7}&\color{blue}{7}&\color{orangered}{24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{4}-7x^{3}+7x^{2}-7x+7 } $ with a remainder of $ \color{red}{ 24 } $.