Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+58x-24 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-7&-17&58&-24\\& & 6& -3& -60& \color{black}{-6} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-20}&\color{blue}{-2}&\color{orangered}{-30} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+58x-24 }{ x-3 } = \color{blue}{2x^{3}-x^{2}-20x-2} \color{red}{~-~} \frac{ \color{red}{ 30 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-7&-17&58&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-7&-17&58&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-7&-17&58&-24\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 6 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -7 }&-17&58&-24\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-7&-17&58&-24\\& & 6& \color{blue}{-3} & & \\ \hline &2&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}3&2&-7&\color{orangered}{ -17 }&58&-24\\& & 6& \color{orangered}{-3} & & \\ \hline &2&-1&\color{orangered}{-20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-7&-17&58&-24\\& & 6& -3& \color{blue}{-60} & \\ \hline &2&-1&\color{blue}{-20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 58 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}3&2&-7&-17&\color{orangered}{ 58 }&-24\\& & 6& -3& \color{orangered}{-60} & \\ \hline &2&-1&-20&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-7&-17&58&-24\\& & 6& -3& -60& \color{blue}{-6} \\ \hline &2&-1&-20&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -30 } $
$$ \begin{array}{c|rrrrr}3&2&-7&-17&58&\color{orangered}{ -24 }\\& & 6& -3& -60& \color{orangered}{-6} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-20}&\color{blue}{-2}&\color{orangered}{-30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-x^{2}-20x-2 } $ with a remainder of $ \color{red}{ -30 } $.