Check if $ x+2 $ is a factor of $ 4x^{3}-5x^{2}-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&4&-5&0&-2\\& & -8& 26& \color{black}{-52} \\ \hline &\color{blue}{4}&\color{blue}{-13}&\color{blue}{26}&\color{orangered}{-54} \end{array} $$Because the remainder $ \left( \color{red}{ -54 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ 4x^{3}-5x^{2}-2$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-5&0&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 4 }&-5&0&-2\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-5&0&-2\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}-2&4&\color{orangered}{ -5 }&0&-2\\& & \color{orangered}{-8} & & \\ \hline &4&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 26 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-5&0&-2\\& & -8& \color{blue}{26} & \\ \hline &4&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 26 } = \color{orangered}{ 26 } $
$$ \begin{array}{c|rrrr}-2&4&-5&\color{orangered}{ 0 }&-2\\& & -8& \color{orangered}{26} & \\ \hline &4&-13&\color{orangered}{26}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 26 } = \color{blue}{ -52 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-5&0&-2\\& & -8& 26& \color{blue}{-52} \\ \hline &4&-13&\color{blue}{26}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -52 \right) } = \color{orangered}{ -54 } $
$$ \begin{array}{c|rrrr}-2&4&-5&0&\color{orangered}{ -2 }\\& & -8& 26& \color{orangered}{-52} \\ \hline &\color{blue}{4}&\color{blue}{-13}&\color{blue}{26}&\color{orangered}{-54} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -54 }\right)$.