Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+58x-24 ) \div ( x-11 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}11&2&-7&-17&58&-24\\& & 22& 165& 1628& \color{black}{18546} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{148}&\color{blue}{1686}&\color{orangered}{18522} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+58x-24 }{ x-11 } = \color{blue}{2x^{3}+15x^{2}+148x+1686} ~+~ \frac{ \color{red}{ 18522 } }{ x-11 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -11 = 0 $ ( $ x = \color{blue}{ 11 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{11}&2&-7&-17&58&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}11&\color{orangered}{ 2 }&-7&-17&58&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 11 } \cdot \color{blue}{ 2 } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{11}&2&-7&-17&58&-24\\& & \color{blue}{22} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 22 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}11&2&\color{orangered}{ -7 }&-17&58&-24\\& & \color{orangered}{22} & & & \\ \hline &2&\color{orangered}{15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 11 } \cdot \color{blue}{ 15 } = \color{blue}{ 165 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{11}&2&-7&-17&58&-24\\& & 22& \color{blue}{165} & & \\ \hline &2&\color{blue}{15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 165 } = \color{orangered}{ 148 } $
$$ \begin{array}{c|rrrrr}11&2&-7&\color{orangered}{ -17 }&58&-24\\& & 22& \color{orangered}{165} & & \\ \hline &2&15&\color{orangered}{148}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 11 } \cdot \color{blue}{ 148 } = \color{blue}{ 1628 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{11}&2&-7&-17&58&-24\\& & 22& 165& \color{blue}{1628} & \\ \hline &2&15&\color{blue}{148}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 58 } + \color{orangered}{ 1628 } = \color{orangered}{ 1686 } $
$$ \begin{array}{c|rrrrr}11&2&-7&-17&\color{orangered}{ 58 }&-24\\& & 22& 165& \color{orangered}{1628} & \\ \hline &2&15&148&\color{orangered}{1686}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 11 } \cdot \color{blue}{ 1686 } = \color{blue}{ 18546 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{11}&2&-7&-17&58&-24\\& & 22& 165& 1628& \color{blue}{18546} \\ \hline &2&15&148&\color{blue}{1686}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 18546 } = \color{orangered}{ 18522 } $
$$ \begin{array}{c|rrrrr}11&2&-7&-17&58&\color{orangered}{ -24 }\\& & 22& 165& 1628& \color{orangered}{18546} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{148}&\color{blue}{1686}&\color{orangered}{18522} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+15x^{2}+148x+1686 } $ with a remainder of $ \color{red}{ 18522 } $.