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