Check if $ 2x+3 $ is a factor of $ 8x^{3}+18x^{2}+17x+12 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&8&18&17&12\\& & -24& 18& \color{black}{-105} \\ \hline &\color{blue}{8}&\color{blue}{-6}&\color{blue}{35}&\color{orangered}{-93} \end{array} $$Because the remainder $ \left( \color{red}{ -93 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ 8x^{3}+18x^{2}+17x+12$.
Explanation
First we need to create a synthetic division table.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&18&17&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 8 }&18&17&12\\& & & & \\ \hline &\color{orangered}{8}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&18&17&12\\& & \color{blue}{-24} & & \\ \hline &\color{blue}{8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-3&8&\color{orangered}{ 18 }&17&12\\& & \color{orangered}{-24} & & \\ \hline &8&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&18&17&12\\& & -24& \color{blue}{18} & \\ \hline &8&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ 18 } = \color{orangered}{ 35 } $
$$ \begin{array}{c|rrrr}-3&8&18&\color{orangered}{ 17 }&12\\& & -24& \color{orangered}{18} & \\ \hline &8&-6&\color{orangered}{35}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 35 } = \color{blue}{ -105 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&18&17&12\\& & -24& 18& \color{blue}{-105} \\ \hline &8&-6&\color{blue}{35}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -105 \right) } = \color{orangered}{ -93 } $
$$ \begin{array}{c|rrrr}-3&8&18&17&\color{orangered}{ 12 }\\& & -24& 18& \color{orangered}{-105} \\ \hline &\color{blue}{8}&\color{blue}{-6}&\color{blue}{35}&\color{orangered}{-93} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -93 }\right)$.