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