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