Check if $ x+2 $ is a factor of $ 3x^{3}+5x^{2}+4x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&5&4&-1\\& & -6& 2& \color{black}{-12} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{6}&\color{orangered}{-13} \end{array} $$Because the remainder $ \left( \color{red}{ -13 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ 3x^{3}+5x^{2}+4x-1$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&4&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&5&4&-1\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&4&-1\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 5 }&4&-1\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&4&-1\\& & -6& \color{blue}{2} & \\ \hline &3&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 2 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-2&3&5&\color{orangered}{ 4 }&-1\\& & -6& \color{orangered}{2} & \\ \hline &3&-1&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&4&-1\\& & -6& 2& \color{blue}{-12} \\ \hline &3&-1&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}-2&3&5&4&\color{orangered}{ -1 }\\& & -6& 2& \color{orangered}{-12} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{6}&\color{orangered}{-13} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -13 }\right)$.