Divide using synthetic division $$ ( 9x^{4}-9x^{3}-58x^{2}+5x+24 ) \div ( x-9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}9&9&-9&-58&5&24\\& & 81& 648& 5310& \color{black}{47835} \\ \hline &\color{blue}{9}&\color{blue}{72}&\color{blue}{590}&\color{blue}{5315}&\color{orangered}{47859} \end{array} $$The solution is:
$$ \frac{ 9x^{4}-9x^{3}-58x^{2}+5x+24 }{ x-9 } = \color{blue}{9x^{3}+72x^{2}+590x+5315} ~+~ \frac{ \color{red}{ 47859 } }{ x-9 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&9&-9&-58&5&24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}9&\color{orangered}{ 9 }&-9&-58&5&24\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 9 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&9&-9&-58&5&24\\& & \color{blue}{81} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 81 } = \color{orangered}{ 72 } $
$$ \begin{array}{c|rrrrr}9&9&\color{orangered}{ -9 }&-58&5&24\\& & \color{orangered}{81} & & & \\ \hline &9&\color{orangered}{72}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 72 } = \color{blue}{ 648 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&9&-9&-58&5&24\\& & 81& \color{blue}{648} & & \\ \hline &9&\color{blue}{72}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -58 } + \color{orangered}{ 648 } = \color{orangered}{ 590 } $
$$ \begin{array}{c|rrrrr}9&9&-9&\color{orangered}{ -58 }&5&24\\& & 81& \color{orangered}{648} & & \\ \hline &9&72&\color{orangered}{590}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 590 } = \color{blue}{ 5310 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&9&-9&-58&5&24\\& & 81& 648& \color{blue}{5310} & \\ \hline &9&72&\color{blue}{590}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 5310 } = \color{orangered}{ 5315 } $
$$ \begin{array}{c|rrrrr}9&9&-9&-58&\color{orangered}{ 5 }&24\\& & 81& 648& \color{orangered}{5310} & \\ \hline &9&72&590&\color{orangered}{5315}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 5315 } = \color{blue}{ 47835 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&9&-9&-58&5&24\\& & 81& 648& 5310& \color{blue}{47835} \\ \hline &9&72&590&\color{blue}{5315}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ 47835 } = \color{orangered}{ 47859 } $
$$ \begin{array}{c|rrrrr}9&9&-9&-58&5&\color{orangered}{ 24 }\\& & 81& 648& 5310& \color{orangered}{47835} \\ \hline &\color{blue}{9}&\color{blue}{72}&\color{blue}{590}&\color{blue}{5315}&\color{orangered}{47859} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{3}+72x^{2}+590x+5315 } $ with a remainder of $ \color{red}{ 47859 } $.