Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+58x-24 ) \div ( x-10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}10&2&-7&-17&58&-24\\& & 20& 130& 1130& \color{black}{11880} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{113}&\color{blue}{1188}&\color{orangered}{11856} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+58x-24 }{ x-10 } = \color{blue}{2x^{3}+13x^{2}+113x+1188} ~+~ \frac{ \color{red}{ 11856 } }{ x-10 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -10 = 0 $ ( $ x = \color{blue}{ 10 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{10}&2&-7&-17&58&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}10&\color{orangered}{ 2 }&-7&-17&58&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 2 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{10}&2&-7&-17&58&-24\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 20 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}10&2&\color{orangered}{ -7 }&-17&58&-24\\& & \color{orangered}{20} & & & \\ \hline &2&\color{orangered}{13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 13 } = \color{blue}{ 130 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{10}&2&-7&-17&58&-24\\& & 20& \color{blue}{130} & & \\ \hline &2&\color{blue}{13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 130 } = \color{orangered}{ 113 } $
$$ \begin{array}{c|rrrrr}10&2&-7&\color{orangered}{ -17 }&58&-24\\& & 20& \color{orangered}{130} & & \\ \hline &2&13&\color{orangered}{113}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 113 } = \color{blue}{ 1130 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{10}&2&-7&-17&58&-24\\& & 20& 130& \color{blue}{1130} & \\ \hline &2&13&\color{blue}{113}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 58 } + \color{orangered}{ 1130 } = \color{orangered}{ 1188 } $
$$ \begin{array}{c|rrrrr}10&2&-7&-17&\color{orangered}{ 58 }&-24\\& & 20& 130& \color{orangered}{1130} & \\ \hline &2&13&113&\color{orangered}{1188}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 1188 } = \color{blue}{ 11880 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{10}&2&-7&-17&58&-24\\& & 20& 130& 1130& \color{blue}{11880} \\ \hline &2&13&113&\color{blue}{1188}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 11880 } = \color{orangered}{ 11856 } $
$$ \begin{array}{c|rrrrr}10&2&-7&-17&58&\color{orangered}{ -24 }\\& & 20& 130& 1130& \color{orangered}{11880} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{113}&\color{blue}{1188}&\color{orangered}{11856} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+13x^{2}+113x+1188 } $ with a remainder of $ \color{red}{ 11856 } $.