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