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