Divide using synthetic division $$ ( 2x^{3}-x^{2}-16x-9 ) \div ( x+9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-9&2&-1&-16&-9\\& & -18& 171& \color{black}{-1395} \\ \hline &\color{blue}{2}&\color{blue}{-19}&\color{blue}{155}&\color{orangered}{-1404} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-x^{2}-16x-9 }{ x+9 } = \color{blue}{2x^{2}-19x+155} \color{red}{~-~} \frac{ \color{red}{ 1404 } }{ x+9 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 9 = 0 $ ( $ x = \color{blue}{ -9 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&2&-1&-16&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-9&\color{orangered}{ 2 }&-1&-16&-9\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 2 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&2&-1&-16&-9\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}-9&2&\color{orangered}{ -1 }&-16&-9\\& & \color{orangered}{-18} & & \\ \hline &2&\color{orangered}{-19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 171 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&2&-1&-16&-9\\& & -18& \color{blue}{171} & \\ \hline &2&\color{blue}{-19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 171 } = \color{orangered}{ 155 } $
$$ \begin{array}{c|rrrr}-9&2&-1&\color{orangered}{ -16 }&-9\\& & -18& \color{orangered}{171} & \\ \hline &2&-19&\color{orangered}{155}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 155 } = \color{blue}{ -1395 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&2&-1&-16&-9\\& & -18& 171& \color{blue}{-1395} \\ \hline &2&-19&\color{blue}{155}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -1395 \right) } = \color{orangered}{ -1404 } $
$$ \begin{array}{c|rrrr}-9&2&-1&-16&\color{orangered}{ -9 }\\& & -18& 171& \color{orangered}{-1395} \\ \hline &\color{blue}{2}&\color{blue}{-19}&\color{blue}{155}&\color{orangered}{-1404} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-19x+155 } $ with a remainder of $ \color{red}{ -1404 } $.