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