Divide using synthetic division $$ ( -x^{4}+5x^{3}+19x^{2}-20 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&-1&5&19&0&-20\\& & 2& -14& -10& \color{black}{20} \\ \hline &\color{blue}{-1}&\color{blue}{7}&\color{blue}{5}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -x^{4}+5x^{3}+19x^{2}-20 }{ x+2 } = \color{blue}{-x^{3}+7x^{2}+5x-10} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&5&19&0&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ -1 }&5&19&0&-20\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&5&19&0&-20\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 2 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}-2&-1&\color{orangered}{ 5 }&19&0&-20\\& & \color{orangered}{2} & & & \\ \hline &-1&\color{orangered}{7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&5&19&0&-20\\& & 2& \color{blue}{-14} & & \\ \hline &-1&\color{blue}{7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-2&-1&5&\color{orangered}{ 19 }&0&-20\\& & 2& \color{orangered}{-14} & & \\ \hline &-1&7&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&5&19&0&-20\\& & 2& -14& \color{blue}{-10} & \\ \hline &-1&7&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&-1&5&19&\color{orangered}{ 0 }&-20\\& & 2& -14& \color{orangered}{-10} & \\ \hline &-1&7&5&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&5&19&0&-20\\& & 2& -14& -10& \color{blue}{20} \\ \hline &-1&7&5&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&-1&5&19&0&\color{orangered}{ -20 }\\& & 2& -14& -10& \color{orangered}{20} \\ \hline &\color{blue}{-1}&\color{blue}{7}&\color{blue}{5}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{3}+7x^{2}+5x-10 } $ with a remainder of $ \color{red}{ 0 } $.