Check if $ x+2 $ is a factor of $ 4x^{4}+2x^{3}-18x-12 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&4&2&0&-18&-12\\& & -8& 12& -24& \color{black}{84} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{12}&\color{blue}{-42}&\color{orangered}{72} \end{array} $$Because the remainder $ \left( \color{red}{ 72 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ 4x^{4}+2x^{3}-18x-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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&2&0&-18&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 4 }&2&0&-18&-12\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&2&0&-18&-12\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&4&\color{orangered}{ 2 }&0&-18&-12\\& & \color{orangered}{-8} & & & \\ \hline &4&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&2&0&-18&-12\\& & -8& \color{blue}{12} & & \\ \hline &4&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-2&4&2&\color{orangered}{ 0 }&-18&-12\\& & -8& \color{orangered}{12} & & \\ \hline &4&-6&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 12 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&2&0&-18&-12\\& & -8& 12& \color{blue}{-24} & \\ \hline &4&-6&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -42 } $
$$ \begin{array}{c|rrrrr}-2&4&2&0&\color{orangered}{ -18 }&-12\\& & -8& 12& \color{orangered}{-24} & \\ \hline &4&-6&12&\color{orangered}{-42}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -42 \right) } = \color{blue}{ 84 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&2&0&-18&-12\\& & -8& 12& -24& \color{blue}{84} \\ \hline &4&-6&12&\color{blue}{-42}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 84 } = \color{orangered}{ 72 } $
$$ \begin{array}{c|rrrrr}-2&4&2&0&-18&\color{orangered}{ -12 }\\& & -8& 12& -24& \color{orangered}{84} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{12}&\color{blue}{-42}&\color{orangered}{72} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 72 }\right)$.