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