Check if $ x+3 $ is a factor of $ 6x^{3}+16x^{2}-7x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&6&16&-7&-1\\& & -18& 6& \color{black}{3} \\ \hline &\color{blue}{6}&\color{blue}{-2}&\color{blue}{-1}&\color{orangered}{2} \end{array} $$Because the remainder $ \left( \color{red}{ 2 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ 6x^{3}+16x^{2}-7x-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 + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&16&-7&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 6 }&16&-7&-1\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&16&-7&-1\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&6&\color{orangered}{ 16 }&-7&-1\\& & \color{orangered}{-18} & & \\ \hline &6&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&16&-7&-1\\& & -18& \color{blue}{6} & \\ \hline &6&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 6 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-3&6&16&\color{orangered}{ -7 }&-1\\& & -18& \color{orangered}{6} & \\ \hline &6&-2&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&16&-7&-1\\& & -18& 6& \color{blue}{3} \\ \hline &6&-2&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-3&6&16&-7&\color{orangered}{ -1 }\\& & -18& 6& \color{orangered}{3} \\ \hline &\color{blue}{6}&\color{blue}{-2}&\color{blue}{-1}&\color{orangered}{2} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 2 }\right)$.