Divide using synthetic division $$ ( 2x^{4}-9x^{3}+9x^{2}+5x-1 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&-9&9&5&-1\\& & -6& 45& -162& \color{black}{471} \\ \hline &\color{blue}{2}&\color{blue}{-15}&\color{blue}{54}&\color{blue}{-157}&\color{orangered}{470} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-9x^{3}+9x^{2}+5x-1 }{ x+3 } = \color{blue}{2x^{3}-15x^{2}+54x-157} ~+~ \frac{ \color{red}{ 470 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-9&9&5&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&-9&9&5&-1\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-9&9&5&-1\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ -9 }&9&5&-1\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{-15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-9&9&5&-1\\& & -6& \color{blue}{45} & & \\ \hline &2&\color{blue}{-15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 45 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrrr}-3&2&-9&\color{orangered}{ 9 }&5&-1\\& & -6& \color{orangered}{45} & & \\ \hline &2&-15&\color{orangered}{54}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 54 } = \color{blue}{ -162 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-9&9&5&-1\\& & -6& 45& \color{blue}{-162} & \\ \hline &2&-15&\color{blue}{54}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -162 \right) } = \color{orangered}{ -157 } $
$$ \begin{array}{c|rrrrr}-3&2&-9&9&\color{orangered}{ 5 }&-1\\& & -6& 45& \color{orangered}{-162} & \\ \hline &2&-15&54&\color{orangered}{-157}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -157 \right) } = \color{blue}{ 471 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-9&9&5&-1\\& & -6& 45& -162& \color{blue}{471} \\ \hline &2&-15&54&\color{blue}{-157}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 471 } = \color{orangered}{ 470 } $
$$ \begin{array}{c|rrrrr}-3&2&-9&9&5&\color{orangered}{ -1 }\\& & -6& 45& -162& \color{orangered}{471} \\ \hline &\color{blue}{2}&\color{blue}{-15}&\color{blue}{54}&\color{blue}{-157}&\color{orangered}{470} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-15x^{2}+54x-157 } $ with a remainder of $ \color{red}{ 470 } $.