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