Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+59x-24 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&2&-7&-17&59&-24\\& & 2& -5& -22& \color{black}{37} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-22}&\color{blue}{37}&\color{orangered}{13} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+59x-24 }{ x-1 } = \color{blue}{2x^{3}-5x^{2}-22x+37} ~+~ \frac{ \color{red}{ 13 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-7&-17&59&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 2 }&-7&-17&59&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-7&-17&59&-24\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&2&\color{orangered}{ -7 }&-17&59&-24\\& & \color{orangered}{2} & & & \\ \hline &2&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-7&-17&59&-24\\& & 2& \color{blue}{-5} & & \\ \hline &2&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrrr}1&2&-7&\color{orangered}{ -17 }&59&-24\\& & 2& \color{orangered}{-5} & & \\ \hline &2&-5&\color{orangered}{-22}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-7&-17&59&-24\\& & 2& -5& \color{blue}{-22} & \\ \hline &2&-5&\color{blue}{-22}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 59 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrrr}1&2&-7&-17&\color{orangered}{ 59 }&-24\\& & 2& -5& \color{orangered}{-22} & \\ \hline &2&-5&-22&\color{orangered}{37}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 37 } = \color{blue}{ 37 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-7&-17&59&-24\\& & 2& -5& -22& \color{blue}{37} \\ \hline &2&-5&-22&\color{blue}{37}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 37 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}1&2&-7&-17&59&\color{orangered}{ -24 }\\& & 2& -5& -22& \color{orangered}{37} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-22}&\color{blue}{37}&\color{orangered}{13} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-5x^{2}-22x+37 } $ with a remainder of $ \color{red}{ 13 } $.