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