Find remainder when $ 6x^{2}+19x+15 $ is divided by $ 2x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}-3&6&19&15\\& & -18& \color{black}{-3} \\ \hline &\color{blue}{6}&\color{blue}{1}&\color{orangered}{12} \end{array} $$The remainder when $ 6x^{2}+19x+15 $ is divided by $ x+3 $ is $ \, \color{red}{ 12 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrr}\color{blue}{-3}&6&19&15\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-3&\color{orangered}{ 6 }&19&15\\& & & \\ \hline &\color{orangered}{6}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&6&19&15\\& & \color{blue}{-18} & \\ \hline &\color{blue}{6}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrr}-3&6&\color{orangered}{ 19 }&15\\& & \color{orangered}{-18} & \\ \hline &6&\color{orangered}{1}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&6&19&15\\& & -18& \color{blue}{-3} \\ \hline &6&\color{blue}{1}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrr}-3&6&19&\color{orangered}{ 15 }\\& & -18& \color{orangered}{-3} \\ \hline &\color{blue}{6}&\color{blue}{1}&\color{orangered}{12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 12 }\right) $.