Divide using synthetic division $$ ( 9x^{4}-9x^{3}-70x^{2}+16x+96 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&9&-9&-70&16&96\\& & 27& 54& -48& \color{black}{-96} \\ \hline &\color{blue}{9}&\color{blue}{18}&\color{blue}{-16}&\color{blue}{-32}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 9x^{4}-9x^{3}-70x^{2}+16x+96 }{ x-3 } = \color{blue}{9x^{3}+18x^{2}-16x-32} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-70&16&96\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 9 }&-9&-70&16&96\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-70&16&96\\& & \color{blue}{27} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 27 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}3&9&\color{orangered}{ -9 }&-70&16&96\\& & \color{orangered}{27} & & & \\ \hline &9&\color{orangered}{18}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 18 } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-70&16&96\\& & 27& \color{blue}{54} & & \\ \hline &9&\color{blue}{18}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -70 } + \color{orangered}{ 54 } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}3&9&-9&\color{orangered}{ -70 }&16&96\\& & 27& \color{orangered}{54} & & \\ \hline &9&18&\color{orangered}{-16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-70&16&96\\& & 27& 54& \color{blue}{-48} & \\ \hline &9&18&\color{blue}{-16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrrr}3&9&-9&-70&\color{orangered}{ 16 }&96\\& & 27& 54& \color{orangered}{-48} & \\ \hline &9&18&-16&\color{orangered}{-32}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -32 \right) } = \color{blue}{ -96 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-70&16&96\\& & 27& 54& -48& \color{blue}{-96} \\ \hline &9&18&-16&\color{blue}{-32}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 96 } + \color{orangered}{ \left( -96 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&9&-9&-70&16&\color{orangered}{ 96 }\\& & 27& 54& -48& \color{orangered}{-96} \\ \hline &\color{blue}{9}&\color{blue}{18}&\color{blue}{-16}&\color{blue}{-32}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{3}+18x^{2}-16x-32 } $ with a remainder of $ \color{red}{ 0 } $.