Divide using synthetic division $$ ( 8x^{3}+14x^{2}+15x+18 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&8&14&15&18\\& & -24& 30& \color{black}{-135} \\ \hline &\color{blue}{8}&\color{blue}{-10}&\color{blue}{45}&\color{orangered}{-117} \end{array} $$The solution is:
$$ \frac{ 8x^{3}+14x^{2}+15x+18 }{ x+3 } = \color{blue}{8x^{2}-10x+45} \color{red}{~-~} \frac{ \color{red}{ 117 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&14&15&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 8 }&14&15&18\\& & & & \\ \hline &\color{orangered}{8}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 8 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&14&15&18\\& & \color{blue}{-24} & & \\ \hline &\color{blue}{8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-3&8&\color{orangered}{ 14 }&15&18\\& & \color{orangered}{-24} & & \\ \hline &8&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&14&15&18\\& & -24& \color{blue}{30} & \\ \hline &8&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 30 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrr}-3&8&14&\color{orangered}{ 15 }&18\\& & -24& \color{orangered}{30} & \\ \hline &8&-10&\color{orangered}{45}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 45 } = \color{blue}{ -135 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&14&15&18\\& & -24& 30& \color{blue}{-135} \\ \hline &8&-10&\color{blue}{45}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -135 \right) } = \color{orangered}{ -117 } $
$$ \begin{array}{c|rrrr}-3&8&14&15&\color{orangered}{ 18 }\\& & -24& 30& \color{orangered}{-135} \\ \hline &\color{blue}{8}&\color{blue}{-10}&\color{blue}{45}&\color{orangered}{-117} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{2}-10x+45 } $ with a remainder of $ \color{red}{ -117 } $.