Divide using synthetic division $$ ( 3x^{3}+x^{2}-5x-6 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&1&-5&-6\\& & -6& 10& \color{black}{-10} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{5}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+x^{2}-5x-6 }{ x+2 } = \color{blue}{3x^{2}-5x+5} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&1&-5&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&1&-5&-6\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&1&-5&-6\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 1 }&-5&-6\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&1&-5&-6\\& & -6& \color{blue}{10} & \\ \hline &3&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 10 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-2&3&1&\color{orangered}{ -5 }&-6\\& & -6& \color{orangered}{10} & \\ \hline &3&-5&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&1&-5&-6\\& & -6& 10& \color{blue}{-10} \\ \hline &3&-5&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-2&3&1&-5&\color{orangered}{ -6 }\\& & -6& 10& \color{orangered}{-10} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{5}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-5x+5 } $ with a remainder of $ \color{red}{ -16 } $.