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