Divide using synthetic division $$ ( 2x^{4}+5x^{3}+3x^{2}+8x+12 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&5&3&8&12\\& & -6& 3& -18& \color{black}{30} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{6}&\color{blue}{-10}&\color{orangered}{42} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+5x^{3}+3x^{2}+8x+12 }{ x+3 } = \color{blue}{2x^{3}-x^{2}+6x-10} ~+~ \frac{ \color{red}{ 42 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&3&8&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&5&3&8&12\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&3&8&12\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ 5 }&3&8&12\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{-1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&3&8&12\\& & -6& \color{blue}{3} & & \\ \hline &2&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 3 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-3&2&5&\color{orangered}{ 3 }&8&12\\& & -6& \color{orangered}{3} & & \\ \hline &2&-1&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&3&8&12\\& & -6& 3& \color{blue}{-18} & \\ \hline &2&-1&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-3&2&5&3&\color{orangered}{ 8 }&12\\& & -6& 3& \color{orangered}{-18} & \\ \hline &2&-1&6&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&3&8&12\\& & -6& 3& -18& \color{blue}{30} \\ \hline &2&-1&6&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 30 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrrrr}-3&2&5&3&8&\color{orangered}{ 12 }\\& & -6& 3& -18& \color{orangered}{30} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{6}&\color{blue}{-10}&\color{orangered}{42} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-x^{2}+6x-10 } $ with a remainder of $ \color{red}{ 42 } $.