Divide using synthetic division $$ ( 3x^{3}+4x^{2}-11x-16 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&4&-11&-16\\& & -6& 4& \color{black}{14} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-7}&\color{orangered}{-2} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+4x^{2}-11x-16 }{ x+2 } = \color{blue}{3x^{2}-2x-7} \color{red}{~-~} \frac{ \color{red}{ 2 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&-11&-16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&4&-11&-16\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&-11&-16\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 4 }&-11&-16\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&-11&-16\\& & -6& \color{blue}{4} & \\ \hline &3&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 4 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-2&3&4&\color{orangered}{ -11 }&-16\\& & -6& \color{orangered}{4} & \\ \hline &3&-2&\color{orangered}{-7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&-11&-16\\& & -6& 4& \color{blue}{14} \\ \hline &3&-2&\color{blue}{-7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 14 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-2&3&4&-11&\color{orangered}{ -16 }\\& & -6& 4& \color{orangered}{14} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-7}&\color{orangered}{-2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-2x-7 } $ with a remainder of $ \color{red}{ -2 } $.