Divide using synthetic division $$ ( 6x^{4}+9x^{3}-11x^{2}+20 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&6&9&-11&0&20\\& & -12& 6& 10& \color{black}{-20} \\ \hline &\color{blue}{6}&\color{blue}{-3}&\color{blue}{-5}&\color{blue}{10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+9x^{3}-11x^{2}+20 }{ x+2 } = \color{blue}{6x^{3}-3x^{2}-5x+10} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&9&-11&0&20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 6 }&9&-11&0&20\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&9&-11&0&20\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-2&6&\color{orangered}{ 9 }&-11&0&20\\& & \color{orangered}{-12} & & & \\ \hline &6&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&9&-11&0&20\\& & -12& \color{blue}{6} & & \\ \hline &6&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 6 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-2&6&9&\color{orangered}{ -11 }&0&20\\& & -12& \color{orangered}{6} & & \\ \hline &6&-3&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&9&-11&0&20\\& & -12& 6& \color{blue}{10} & \\ \hline &6&-3&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-2&6&9&-11&\color{orangered}{ 0 }&20\\& & -12& 6& \color{orangered}{10} & \\ \hline &6&-3&-5&\color{orangered}{10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 10 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&9&-11&0&20\\& & -12& 6& 10& \color{blue}{-20} \\ \hline &6&-3&-5&\color{blue}{10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&6&9&-11&0&\color{orangered}{ 20 }\\& & -12& 6& 10& \color{orangered}{-20} \\ \hline &\color{blue}{6}&\color{blue}{-3}&\color{blue}{-5}&\color{blue}{10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}-3x^{2}-5x+10 } $ with a remainder of $ \color{red}{ 0 } $.