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