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