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