Divide using synthetic division $$ ( x^{4}+5x^{3}-2x^{2}+6x-2 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&1&5&-2&6&-2\\& & -1& -4& 6& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-6}&\color{blue}{12}&\color{orangered}{-14} \end{array} $$The solution is:
$$ \frac{ x^{4}+5x^{3}-2x^{2}+6x-2 }{ x+1 } = \color{blue}{x^{3}+4x^{2}-6x+12} \color{red}{~-~} \frac{ \color{red}{ 14 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&5&-2&6&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 1 }&5&-2&6&-2\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&5&-2&6&-2\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-1&1&\color{orangered}{ 5 }&-2&6&-2\\& & \color{orangered}{-1} & & & \\ \hline &1&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&5&-2&6&-2\\& & -1& \color{blue}{-4} & & \\ \hline &1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-1&1&5&\color{orangered}{ -2 }&6&-2\\& & -1& \color{orangered}{-4} & & \\ \hline &1&4&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&5&-2&6&-2\\& & -1& -4& \color{blue}{6} & \\ \hline &1&4&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 6 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-1&1&5&-2&\color{orangered}{ 6 }&-2\\& & -1& -4& \color{orangered}{6} & \\ \hline &1&4&-6&\color{orangered}{12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 12 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&5&-2&6&-2\\& & -1& -4& 6& \color{blue}{-12} \\ \hline &1&4&-6&\color{blue}{12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrr}-1&1&5&-2&6&\color{orangered}{ -2 }\\& & -1& -4& 6& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-6}&\color{blue}{12}&\color{orangered}{-14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+4x^{2}-6x+12 } $ with a remainder of $ \color{red}{ -14 } $.