Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+58x-24 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&2&-7&-17&58&-24\\& & 10& 15& -10& \color{black}{240} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-2}&\color{blue}{48}&\color{orangered}{216} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+58x-24 }{ x-5 } = \color{blue}{2x^{3}+3x^{2}-2x+48} ~+~ \frac{ \color{red}{ 216 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-7&-17&58&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 2 }&-7&-17&58&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-7&-17&58&-24\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 10 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}5&2&\color{orangered}{ -7 }&-17&58&-24\\& & \color{orangered}{10} & & & \\ \hline &2&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-7&-17&58&-24\\& & 10& \color{blue}{15} & & \\ \hline &2&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 15 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&2&-7&\color{orangered}{ -17 }&58&-24\\& & 10& \color{orangered}{15} & & \\ \hline &2&3&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-7&-17&58&-24\\& & 10& 15& \color{blue}{-10} & \\ \hline &2&3&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 58 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrr}5&2&-7&-17&\color{orangered}{ 58 }&-24\\& & 10& 15& \color{orangered}{-10} & \\ \hline &2&3&-2&\color{orangered}{48}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 48 } = \color{blue}{ 240 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-7&-17&58&-24\\& & 10& 15& -10& \color{blue}{240} \\ \hline &2&3&-2&\color{blue}{48}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 240 } = \color{orangered}{ 216 } $
$$ \begin{array}{c|rrrrr}5&2&-7&-17&58&\color{orangered}{ -24 }\\& & 10& 15& -10& \color{orangered}{240} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-2}&\color{blue}{48}&\color{orangered}{216} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+3x^{2}-2x+48 } $ with a remainder of $ \color{red}{ 216 } $.