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