Divide using synthetic division $$ ( x^{5}+6x^{4}-11x^{3}-84x^{2}+28x+240 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&6&-11&-84&28&240\\& & -2& -8& 38& 92& \color{black}{-240} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-19}&\color{blue}{-46}&\color{blue}{120}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}+6x^{4}-11x^{3}-84x^{2}+28x+240 }{ x+2 } = \color{blue}{x^{4}+4x^{3}-19x^{2}-46x+120} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&6&-11&-84&28&240\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&6&-11&-84&28&240\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&6&-11&-84&28&240\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ 6 }&-11&-84&28&240\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&6&-11&-84&28&240\\& & -2& \color{blue}{-8} & & & \\ \hline &1&\color{blue}{4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrrrr}-2&1&6&\color{orangered}{ -11 }&-84&28&240\\& & -2& \color{orangered}{-8} & & & \\ \hline &1&4&\color{orangered}{-19}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 38 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&6&-11&-84&28&240\\& & -2& -8& \color{blue}{38} & & \\ \hline &1&4&\color{blue}{-19}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -84 } + \color{orangered}{ 38 } = \color{orangered}{ -46 } $
$$ \begin{array}{c|rrrrrr}-2&1&6&-11&\color{orangered}{ -84 }&28&240\\& & -2& -8& \color{orangered}{38} & & \\ \hline &1&4&-19&\color{orangered}{-46}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -46 \right) } = \color{blue}{ 92 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&6&-11&-84&28&240\\& & -2& -8& 38& \color{blue}{92} & \\ \hline &1&4&-19&\color{blue}{-46}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ 92 } = \color{orangered}{ 120 } $
$$ \begin{array}{c|rrrrrr}-2&1&6&-11&-84&\color{orangered}{ 28 }&240\\& & -2& -8& 38& \color{orangered}{92} & \\ \hline &1&4&-19&-46&\color{orangered}{120}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 120 } = \color{blue}{ -240 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&6&-11&-84&28&240\\& & -2& -8& 38& 92& \color{blue}{-240} \\ \hline &1&4&-19&-46&\color{blue}{120}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 240 } + \color{orangered}{ \left( -240 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-2&1&6&-11&-84&28&\color{orangered}{ 240 }\\& & -2& -8& 38& 92& \color{orangered}{-240} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-19}&\color{blue}{-46}&\color{blue}{120}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+4x^{3}-19x^{2}-46x+120 } $ with a remainder of $ \color{red}{ 0 } $.