Divide using synthetic division $$ ( 9x^{4}-9x^{3}-58x^{2}+5x+24 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&9&-9&-58&5&24\\& & 9& 0& -58& \color{black}{-53} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{-58}&\color{blue}{-53}&\color{orangered}{-29} \end{array} $$The solution is:
$$ \frac{ 9x^{4}-9x^{3}-58x^{2}+5x+24 }{ x-1 } = \color{blue}{9x^{3}-58x-53} \color{red}{~-~} \frac{ \color{red}{ 29 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-9&-58&5&24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 9 }&-9&-58&5&24\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-9&-58&5&24\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 9 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&9&\color{orangered}{ -9 }&-58&5&24\\& & \color{orangered}{9} & & & \\ \hline &9&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-9&-58&5&24\\& & 9& \color{blue}{0} & & \\ \hline &9&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -58 } + \color{orangered}{ 0 } = \color{orangered}{ -58 } $
$$ \begin{array}{c|rrrrr}1&9&-9&\color{orangered}{ -58 }&5&24\\& & 9& \color{orangered}{0} & & \\ \hline &9&0&\color{orangered}{-58}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -58 \right) } = \color{blue}{ -58 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-9&-58&5&24\\& & 9& 0& \color{blue}{-58} & \\ \hline &9&0&\color{blue}{-58}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -58 \right) } = \color{orangered}{ -53 } $
$$ \begin{array}{c|rrrrr}1&9&-9&-58&\color{orangered}{ 5 }&24\\& & 9& 0& \color{orangered}{-58} & \\ \hline &9&0&-58&\color{orangered}{-53}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -53 \right) } = \color{blue}{ -53 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-9&-58&5&24\\& & 9& 0& -58& \color{blue}{-53} \\ \hline &9&0&-58&\color{blue}{-53}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -53 \right) } = \color{orangered}{ -29 } $
$$ \begin{array}{c|rrrrr}1&9&-9&-58&5&\color{orangered}{ 24 }\\& & 9& 0& -58& \color{orangered}{-53} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{-58}&\color{blue}{-53}&\color{orangered}{-29} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{3}-58x-53 } $ with a remainder of $ \color{red}{ -29 } $.