Divide using synthetic division $$ ( 2x^{3}+9x^{2}+13x+6 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&9&13&6\\& & -6& -9& \color{black}{-12} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+9x^{2}+13x+6 }{ x+3 } = \color{blue}{2x^{2}+3x+4} \color{red}{~-~} \frac{ \color{red}{ 6 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&9&13&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&9&13&6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&9&13&6\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ 9 }&13&6\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&9&13&6\\& & -6& \color{blue}{-9} & \\ \hline &2&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-3&2&9&\color{orangered}{ 13 }&6\\& & -6& \color{orangered}{-9} & \\ \hline &2&3&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&9&13&6\\& & -6& -9& \color{blue}{-12} \\ \hline &2&3&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-3&2&9&13&\color{orangered}{ 6 }\\& & -6& -9& \color{orangered}{-12} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+3x+4 } $ with a remainder of $ \color{red}{ -6 } $.