Divide using synthetic division $$ ( 6x^{4}+11x^{3}+7x^{2}+11x-14 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&6&11&7&11&-14\\& & -12& 2& -18& \color{black}{14} \\ \hline &\color{blue}{6}&\color{blue}{-1}&\color{blue}{9}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+11x^{3}+7x^{2}+11x-14 }{ x+2 } = \color{blue}{6x^{3}-x^{2}+9x-7} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&11&7&11&-14\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 6 }&11&7&11&-14\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&11&7&11&-14\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-2&6&\color{orangered}{ 11 }&7&11&-14\\& & \color{orangered}{-12} & & & \\ \hline &6&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&11&7&11&-14\\& & -12& \color{blue}{2} & & \\ \hline &6&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 2 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-2&6&11&\color{orangered}{ 7 }&11&-14\\& & -12& \color{orangered}{2} & & \\ \hline &6&-1&\color{orangered}{9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 9 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&11&7&11&-14\\& & -12& 2& \color{blue}{-18} & \\ \hline &6&-1&\color{blue}{9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-2&6&11&7&\color{orangered}{ 11 }&-14\\& & -12& 2& \color{orangered}{-18} & \\ \hline &6&-1&9&\color{orangered}{-7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&11&7&11&-14\\& & -12& 2& -18& \color{blue}{14} \\ \hline &6&-1&9&\color{blue}{-7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 14 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&6&11&7&11&\color{orangered}{ -14 }\\& & -12& 2& -18& \color{orangered}{14} \\ \hline &\color{blue}{6}&\color{blue}{-1}&\color{blue}{9}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}-x^{2}+9x-7 } $ with a remainder of $ \color{red}{ 0 } $.