Divide using synthetic division $$ ( 2x^{3}-9x^{2}-16x-2 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&2&-9&-16&-2\\& & -2& 11& \color{black}{5} \\ \hline &\color{blue}{2}&\color{blue}{-11}&\color{blue}{-5}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-9x^{2}-16x-2 }{ x+1 } = \color{blue}{2x^{2}-11x-5} ~+~ \frac{ \color{red}{ 3 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-9&-16&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 2 }&-9&-16&-2\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-9&-16&-2\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-1&2&\color{orangered}{ -9 }&-16&-2\\& & \color{orangered}{-2} & & \\ \hline &2&\color{orangered}{-11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-9&-16&-2\\& & -2& \color{blue}{11} & \\ \hline &2&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 11 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-1&2&-9&\color{orangered}{ -16 }&-2\\& & -2& \color{orangered}{11} & \\ \hline &2&-11&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-9&-16&-2\\& & -2& 11& \color{blue}{5} \\ \hline &2&-11&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 5 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-1&2&-9&-16&\color{orangered}{ -2 }\\& & -2& 11& \color{orangered}{5} \\ \hline &\color{blue}{2}&\color{blue}{-11}&\color{blue}{-5}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-11x-5 } $ with a remainder of $ \color{red}{ 3 } $.