Divide using synthetic division $$ ( 9x^{4}-9x^{3}-58x^{2}+5x+24 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&9&-9&-58&5&24\\& & 27& 54& -12& \color{black}{-21} \\ \hline &\color{blue}{9}&\color{blue}{18}&\color{blue}{-4}&\color{blue}{-7}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 9x^{4}-9x^{3}-58x^{2}+5x+24 }{ x-3 } = \color{blue}{9x^{3}+18x^{2}-4x-7} ~+~ \frac{ \color{red}{ 3 } }{ x-3 } $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-58&5&24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 9 }&-9&-58&5&24\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-58&5&24\\& & \color{blue}{27} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 27 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}3&9&\color{orangered}{ -9 }&-58&5&24\\& & \color{orangered}{27} & & & \\ \hline &9&\color{orangered}{18}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-58&5&24\\& & 27& \color{blue}{54} & & \\ \hline &9&\color{blue}{18}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -58 } + \color{orangered}{ 54 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}3&9&-9&\color{orangered}{ -58 }&5&24\\& & 27& \color{orangered}{54} & & \\ \hline &9&18&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-58&5&24\\& & 27& 54& \color{blue}{-12} & \\ \hline &9&18&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}3&9&-9&-58&\color{orangered}{ 5 }&24\\& & 27& 54& \color{orangered}{-12} & \\ \hline &9&18&-4&\color{orangered}{-7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&9&-9&-58&5&24\\& & 27& 54& -12& \color{blue}{-21} \\ \hline &9&18&-4&\color{blue}{-7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&9&-9&-58&5&\color{orangered}{ 24 }\\& & 27& 54& -12& \color{orangered}{-21} \\ \hline &\color{blue}{9}&\color{blue}{18}&\color{blue}{-4}&\color{blue}{-7}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{3}+18x^{2}-4x-7 } $ with a remainder of $ \color{red}{ 3 } $.