Check if $ x-2 $ is a factor of $ 3x^{3}-13x^{2}+8x+12 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&3&-13&8&12\\& & 6& -14& \color{black}{-12} \\ \hline &\color{blue}{3}&\color{blue}{-7}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-2 $ is a factor of the $ 3x^{3}-13x^{2}+8x+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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-13&8&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 3 }&-13&8&12\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-13&8&12\\& & \color{blue}{6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 6 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}2&3&\color{orangered}{ -13 }&8&12\\& & \color{orangered}{6} & & \\ \hline &3&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-13&8&12\\& & 6& \color{blue}{-14} & \\ \hline &3&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}2&3&-13&\color{orangered}{ 8 }&12\\& & 6& \color{orangered}{-14} & \\ \hline &3&-7&\color{orangered}{-6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-13&8&12\\& & 6& -14& \color{blue}{-12} \\ \hline &3&-7&\color{blue}{-6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&3&-13&8&\color{orangered}{ 12 }\\& & 6& -14& \color{orangered}{-12} \\ \hline &\color{blue}{3}&\color{blue}{-7}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.