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