Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+59x-24 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&2&-7&-17&59&-24\\& & 12& 30& 78& \color{black}{822} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{13}&\color{blue}{137}&\color{orangered}{798} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+59x-24 }{ x-6 } = \color{blue}{2x^{3}+5x^{2}+13x+137} ~+~ \frac{ \color{red}{ 798 } }{ x-6 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-7&-17&59&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}6&\color{orangered}{ 2 }&-7&-17&59&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-7&-17&59&-24\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 12 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}6&2&\color{orangered}{ -7 }&-17&59&-24\\& & \color{orangered}{12} & & & \\ \hline &2&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 5 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-7&-17&59&-24\\& & 12& \color{blue}{30} & & \\ \hline &2&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 30 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}6&2&-7&\color{orangered}{ -17 }&59&-24\\& & 12& \color{orangered}{30} & & \\ \hline &2&5&\color{orangered}{13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 13 } = \color{blue}{ 78 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-7&-17&59&-24\\& & 12& 30& \color{blue}{78} & \\ \hline &2&5&\color{blue}{13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 59 } + \color{orangered}{ 78 } = \color{orangered}{ 137 } $
$$ \begin{array}{c|rrrrr}6&2&-7&-17&\color{orangered}{ 59 }&-24\\& & 12& 30& \color{orangered}{78} & \\ \hline &2&5&13&\color{orangered}{137}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 137 } = \color{blue}{ 822 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-7&-17&59&-24\\& & 12& 30& 78& \color{blue}{822} \\ \hline &2&5&13&\color{blue}{137}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 822 } = \color{orangered}{ 798 } $
$$ \begin{array}{c|rrrrr}6&2&-7&-17&59&\color{orangered}{ -24 }\\& & 12& 30& 78& \color{orangered}{822} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{13}&\color{blue}{137}&\color{orangered}{798} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+5x^{2}+13x+137 } $ with a remainder of $ \color{red}{ 798 } $.