Divide using synthetic division $$ ( -x^{3}-7x^{2}+8x+4 ) \div ( -2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&-1&-7&8&4\\& & 3& 12& \color{black}{-60} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{20}&\color{orangered}{-56} \end{array} $$The solution is:
$$ \frac{ -x^{3}-7x^{2}+8x+4 }{ x+3 } = \color{blue}{-x^{2}-4x+20} \color{red}{~-~} \frac{ \color{red}{ 56 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&-7&8&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ -1 }&-7&8&4\\& & & & \\ \hline &\color{orangered}{-1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&-7&8&4\\& & \color{blue}{3} & & \\ \hline &\color{blue}{-1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 3 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-3&-1&\color{orangered}{ -7 }&8&4\\& & \color{orangered}{3} & & \\ \hline &-1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&-7&8&4\\& & 3& \color{blue}{12} & \\ \hline &-1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 12 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}-3&-1&-7&\color{orangered}{ 8 }&4\\& & 3& \color{orangered}{12} & \\ \hline &-1&-4&\color{orangered}{20}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 20 } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-1&-7&8&4\\& & 3& 12& \color{blue}{-60} \\ \hline &-1&-4&\color{blue}{20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -56 } $
$$ \begin{array}{c|rrrr}-3&-1&-7&8&\color{orangered}{ 4 }\\& & 3& 12& \color{orangered}{-60} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{20}&\color{orangered}{-56} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{2}-4x+20 } $ with a remainder of $ \color{red}{ -56 } $.