Check if $ 2x+3 $ is a factor of $ x^{4}-6x^{3}+12x^{2}+6x-13 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-6&12&6&-13\\& & -3& 27& -117& \color{black}{333} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{39}&\color{blue}{-111}&\color{orangered}{320} \end{array} $$Because the remainder $ \left( \color{red}{ 320 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ x^{4}-6x^{3}+12x^{2}+6x-13$.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&12&6&-13\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-6&12&6&-13\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&12&6&-13\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -6 }&12&6&-13\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&12&6&-13\\& & -3& \color{blue}{27} & & \\ \hline &1&\color{blue}{-9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 27 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrrr}-3&1&-6&\color{orangered}{ 12 }&6&-13\\& & -3& \color{orangered}{27} & & \\ \hline &1&-9&\color{orangered}{39}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 39 } = \color{blue}{ -117 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&12&6&-13\\& & -3& 27& \color{blue}{-117} & \\ \hline &1&-9&\color{blue}{39}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -117 \right) } = \color{orangered}{ -111 } $
$$ \begin{array}{c|rrrrr}-3&1&-6&12&\color{orangered}{ 6 }&-13\\& & -3& 27& \color{orangered}{-117} & \\ \hline &1&-9&39&\color{orangered}{-111}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -111 \right) } = \color{blue}{ 333 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&12&6&-13\\& & -3& 27& -117& \color{blue}{333} \\ \hline &1&-9&39&\color{blue}{-111}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 333 } = \color{orangered}{ 320 } $
$$ \begin{array}{c|rrrrr}-3&1&-6&12&6&\color{orangered}{ -13 }\\& & -3& 27& -117& \color{orangered}{333} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{39}&\color{blue}{-111}&\color{orangered}{320} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 320 }\right)$.