Divide using synthetic division $$ ( 3x^{3}+11x^{2}-10x-7 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&11&-10&-7\\& & -6& -10& \color{black}{40} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{-20}&\color{orangered}{33} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+11x^{2}-10x-7 }{ x+2 } = \color{blue}{3x^{2}+5x-20} ~+~ \frac{ \color{red}{ 33 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&-10&-7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&11&-10&-7\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&-10&-7\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 11 }&-10&-7\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&-10&-7\\& & -6& \color{blue}{-10} & \\ \hline &3&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrr}-2&3&11&\color{orangered}{ -10 }&-7\\& & -6& \color{orangered}{-10} & \\ \hline &3&5&\color{orangered}{-20}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&-10&-7\\& & -6& -10& \color{blue}{40} \\ \hline &3&5&\color{blue}{-20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 40 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrr}-2&3&11&-10&\color{orangered}{ -7 }\\& & -6& -10& \color{orangered}{40} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{-20}&\color{orangered}{33} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+5x-20 } $ with a remainder of $ \color{red}{ 33 } $.