Divide using synthetic division $$ ( 2x^{3}+6x^{2}-11x-40 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&2&6&-11&-40\\& & -10& 20& \color{black}{-45} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{9}&\color{orangered}{-85} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+6x^{2}-11x-40 }{ x+5 } = \color{blue}{2x^{2}-4x+9} \color{red}{~-~} \frac{ \color{red}{ 85 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&6&-11&-40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 2 }&6&-11&-40\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&6&-11&-40\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-5&2&\color{orangered}{ 6 }&-11&-40\\& & \color{orangered}{-10} & & \\ \hline &2&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&6&-11&-40\\& & -10& \color{blue}{20} & \\ \hline &2&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 20 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}-5&2&6&\color{orangered}{ -11 }&-40\\& & -10& \color{orangered}{20} & \\ \hline &2&-4&\color{orangered}{9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 9 } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&6&-11&-40\\& & -10& 20& \color{blue}{-45} \\ \hline &2&-4&\color{blue}{9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ -85 } $
$$ \begin{array}{c|rrrr}-5&2&6&-11&\color{orangered}{ -40 }\\& & -10& 20& \color{orangered}{-45} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{9}&\color{orangered}{-85} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-4x+9 } $ with a remainder of $ \color{red}{ -85 } $.