Divide using synthetic division $$ ( 9x^{3}-18x^{2}-16x+32 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&9&-18&-16&32\\& & 18& 0& \color{black}{-32} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{-16}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 9x^{3}-18x^{2}-16x+32 }{ x-2 } = \color{blue}{9x^{2}-16} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&-18&-16&32\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 9 }&-18&-16&32\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&-18&-16&32\\& & \color{blue}{18} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 18 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&9&\color{orangered}{ -18 }&-16&32\\& & \color{orangered}{18} & & \\ \hline &9&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&-18&-16&32\\& & 18& \color{blue}{0} & \\ \hline &9&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 0 } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}2&9&-18&\color{orangered}{ -16 }&32\\& & 18& \color{orangered}{0} & \\ \hline &9&0&\color{orangered}{-16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&-18&-16&32\\& & 18& 0& \color{blue}{-32} \\ \hline &9&0&\color{blue}{-16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&9&-18&-16&\color{orangered}{ 32 }\\& & 18& 0& \color{orangered}{-32} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{-16}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}-16 } $ with a remainder of $ \color{red}{ 0 } $.