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