Divide using synthetic division $$ ( -7x^{4}-11x^{3}+8x^{2}+10x+4 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&-7&-11&8&10&4\\& & 14& -6& -4& \color{black}{-12} \\ \hline &\color{blue}{-7}&\color{blue}{3}&\color{blue}{2}&\color{blue}{6}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ -7x^{4}-11x^{3}+8x^{2}+10x+4 }{ x+2 } = \color{blue}{-7x^{3}+3x^{2}+2x+6} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-7&-11&8&10&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ -7 }&-11&8&10&4\\& & & & & \\ \hline &\color{orangered}{-7}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-7&-11&8&10&4\\& & \color{blue}{14} & & & \\ \hline &\color{blue}{-7}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 14 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-2&-7&\color{orangered}{ -11 }&8&10&4\\& & \color{orangered}{14} & & & \\ \hline &-7&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-7&-11&8&10&4\\& & 14& \color{blue}{-6} & & \\ \hline &-7&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-2&-7&-11&\color{orangered}{ 8 }&10&4\\& & 14& \color{orangered}{-6} & & \\ \hline &-7&3&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-7&-11&8&10&4\\& & 14& -6& \color{blue}{-4} & \\ \hline &-7&3&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&-7&-11&8&\color{orangered}{ 10 }&4\\& & 14& -6& \color{orangered}{-4} & \\ \hline &-7&3&2&\color{orangered}{6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-7&-11&8&10&4\\& & 14& -6& -4& \color{blue}{-12} \\ \hline &-7&3&2&\color{blue}{6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&-7&-11&8&10&\color{orangered}{ 4 }\\& & 14& -6& -4& \color{orangered}{-12} \\ \hline &\color{blue}{-7}&\color{blue}{3}&\color{blue}{2}&\color{blue}{6}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -7x^{3}+3x^{2}+2x+6 } $ with a remainder of $ \color{red}{ -8 } $.