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