Divide using synthetic division $$ ( 2x^{3}+5x^{2}+8x+3 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&5&8&3\\& & -6& 3& \color{black}{-33} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{11}&\color{orangered}{-30} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+5x^{2}+8x+3 }{ x+3 } = \color{blue}{2x^{2}-x+11} \color{red}{~-~} \frac{ \color{red}{ 30 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&5&8&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&5&8&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&5&8&3\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ 5 }&8&3\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-1}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&5&8&3\\& & -6& \color{blue}{3} & \\ \hline &2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 3 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}-3&2&5&\color{orangered}{ 8 }&3\\& & -6& \color{orangered}{3} & \\ \hline &2&-1&\color{orangered}{11}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 11 } = \color{blue}{ -33 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&5&8&3\\& & -6& 3& \color{blue}{-33} \\ \hline &2&-1&\color{blue}{11}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -33 \right) } = \color{orangered}{ -30 } $
$$ \begin{array}{c|rrrr}-3&2&5&8&\color{orangered}{ 3 }\\& & -6& 3& \color{orangered}{-33} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{11}&\color{orangered}{-30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-x+11 } $ with a remainder of $ \color{red}{ -30 } $.