Find remainder when $ 12x^{3}-x^{2}-7x+2 $ is divided by $ 3x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&12&-1&-7&2\\& & 12& 11& \color{black}{4} \\ \hline &\color{blue}{12}&\color{blue}{11}&\color{blue}{4}&\color{orangered}{6} \end{array} $$The remainder when $ 12x^{3}-x^{2}-7x+2 $ is divided by $ x-1 $ is $ \, \color{red}{ 6 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&-1&-7&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 12 }&-1&-7&2\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 12 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&-1&-7&2\\& & \color{blue}{12} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 12 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}1&12&\color{orangered}{ -1 }&-7&2\\& & \color{orangered}{12} & & \\ \hline &12&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 11 } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&-1&-7&2\\& & 12& \color{blue}{11} & \\ \hline &12&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 11 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}1&12&-1&\color{orangered}{ -7 }&2\\& & 12& \color{orangered}{11} & \\ \hline &12&11&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&-1&-7&2\\& & 12& 11& \color{blue}{4} \\ \hline &12&11&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 4 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}1&12&-1&-7&\color{orangered}{ 2 }\\& & 12& 11& \color{orangered}{4} \\ \hline &\color{blue}{12}&\color{blue}{11}&\color{blue}{4}&\color{orangered}{6} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 6 }\right) $.