Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+59x-24 ) \div ( x-12 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}12&2&-7&-17&59&-24\\& & 24& 204& 2244& \color{black}{27636} \\ \hline &\color{blue}{2}&\color{blue}{17}&\color{blue}{187}&\color{blue}{2303}&\color{orangered}{27612} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+59x-24 }{ x-12 } = \color{blue}{2x^{3}+17x^{2}+187x+2303} ~+~ \frac{ \color{red}{ 27612 } }{ x-12 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -12 = 0 $ ( $ x = \color{blue}{ 12 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&-7&-17&59&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}12&\color{orangered}{ 2 }&-7&-17&59&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 2 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&-7&-17&59&-24\\& & \color{blue}{24} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 24 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}12&2&\color{orangered}{ -7 }&-17&59&-24\\& & \color{orangered}{24} & & & \\ \hline &2&\color{orangered}{17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 17 } = \color{blue}{ 204 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&-7&-17&59&-24\\& & 24& \color{blue}{204} & & \\ \hline &2&\color{blue}{17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 204 } = \color{orangered}{ 187 } $
$$ \begin{array}{c|rrrrr}12&2&-7&\color{orangered}{ -17 }&59&-24\\& & 24& \color{orangered}{204} & & \\ \hline &2&17&\color{orangered}{187}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 187 } = \color{blue}{ 2244 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&-7&-17&59&-24\\& & 24& 204& \color{blue}{2244} & \\ \hline &2&17&\color{blue}{187}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 59 } + \color{orangered}{ 2244 } = \color{orangered}{ 2303 } $
$$ \begin{array}{c|rrrrr}12&2&-7&-17&\color{orangered}{ 59 }&-24\\& & 24& 204& \color{orangered}{2244} & \\ \hline &2&17&187&\color{orangered}{2303}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 2303 } = \color{blue}{ 27636 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&-7&-17&59&-24\\& & 24& 204& 2244& \color{blue}{27636} \\ \hline &2&17&187&\color{blue}{2303}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 27636 } = \color{orangered}{ 27612 } $
$$ \begin{array}{c|rrrrr}12&2&-7&-17&59&\color{orangered}{ -24 }\\& & 24& 204& 2244& \color{orangered}{27636} \\ \hline &\color{blue}{2}&\color{blue}{17}&\color{blue}{187}&\color{blue}{2303}&\color{orangered}{27612} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+17x^{2}+187x+2303 } $ with a remainder of $ \color{red}{ 27612 } $.