Divide using synthetic division $$ ( 2x^{4}-3x^{3}+4x^{2}+17x+7 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&-3&4&17&7\\& & -6& 27& -93& \color{black}{228} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{31}&\color{blue}{-76}&\color{orangered}{235} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-3x^{3}+4x^{2}+17x+7 }{ x+3 } = \color{blue}{2x^{3}-9x^{2}+31x-76} ~+~ \frac{ \color{red}{ 235 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-3&4&17&7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&-3&4&17&7\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-3&4&17&7\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ -3 }&4&17&7\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{-9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-3&4&17&7\\& & -6& \color{blue}{27} & & \\ \hline &2&\color{blue}{-9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 27 } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrrr}-3&2&-3&\color{orangered}{ 4 }&17&7\\& & -6& \color{orangered}{27} & & \\ \hline &2&-9&\color{orangered}{31}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 31 } = \color{blue}{ -93 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-3&4&17&7\\& & -6& 27& \color{blue}{-93} & \\ \hline &2&-9&\color{blue}{31}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -93 \right) } = \color{orangered}{ -76 } $
$$ \begin{array}{c|rrrrr}-3&2&-3&4&\color{orangered}{ 17 }&7\\& & -6& 27& \color{orangered}{-93} & \\ \hline &2&-9&31&\color{orangered}{-76}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -76 \right) } = \color{blue}{ 228 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-3&4&17&7\\& & -6& 27& -93& \color{blue}{228} \\ \hline &2&-9&31&\color{blue}{-76}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 228 } = \color{orangered}{ 235 } $
$$ \begin{array}{c|rrrrr}-3&2&-3&4&17&\color{orangered}{ 7 }\\& & -6& 27& -93& \color{orangered}{228} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{31}&\color{blue}{-76}&\color{orangered}{235} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-9x^{2}+31x-76 } $ with a remainder of $ \color{red}{ 235 } $.