Check if $ x-1 $ is a factor of $ 3x^{4}-16x^{3}+21x^{2}+4x-12 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&3&-16&21&4&-12\\& & 3& -13& 8& \color{black}{12} \\ \hline &\color{blue}{3}&\color{blue}{-13}&\color{blue}{8}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-1 $ is a factor of the $ 3x^{4}-16x^{3}+21x^{2}+4x-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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&21&4&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 3 }&-16&21&4&-12\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&21&4&-12\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 3 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}1&3&\color{orangered}{ -16 }&21&4&-12\\& & \color{orangered}{3} & & & \\ \hline &3&\color{orangered}{-13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&21&4&-12\\& & 3& \color{blue}{-13} & & \\ \hline &3&\color{blue}{-13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}1&3&-16&\color{orangered}{ 21 }&4&-12\\& & 3& \color{orangered}{-13} & & \\ \hline &3&-13&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&21&4&-12\\& & 3& -13& \color{blue}{8} & \\ \hline &3&-13&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 8 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}1&3&-16&21&\color{orangered}{ 4 }&-12\\& & 3& -13& \color{orangered}{8} & \\ \hline &3&-13&8&\color{orangered}{12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 12 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&21&4&-12\\& & 3& -13& 8& \color{blue}{12} \\ \hline &3&-13&8&\color{blue}{12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&3&-16&21&4&\color{orangered}{ -12 }\\& & 3& -13& 8& \color{orangered}{12} \\ \hline &\color{blue}{3}&\color{blue}{-13}&\color{blue}{8}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.