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