Divide using synthetic division $$ ( 12x^{3}+2x^{2}-5x+1 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&12&2&-5&1\\& & 12& 14& \color{black}{9} \\ \hline &\color{blue}{12}&\color{blue}{14}&\color{blue}{9}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ 12x^{3}+2x^{2}-5x+1 }{ x-1 } = \color{blue}{12x^{2}+14x+9} ~+~ \frac{ \color{red}{ 10 } }{ x-1 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&2&-5&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 12 }&2&-5&1\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&2&-5&1\\& & \color{blue}{12} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 12 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}1&12&\color{orangered}{ 2 }&-5&1\\& & \color{orangered}{12} & & \\ \hline &12&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 14 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&2&-5&1\\& & 12& \color{blue}{14} & \\ \hline &12&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 14 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}1&12&2&\color{orangered}{ -5 }&1\\& & 12& \color{orangered}{14} & \\ \hline &12&14&\color{orangered}{9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&2&-5&1\\& & 12& 14& \color{blue}{9} \\ \hline &12&14&\color{blue}{9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 9 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}1&12&2&-5&\color{orangered}{ 1 }\\& & 12& 14& \color{orangered}{9} \\ \hline &\color{blue}{12}&\color{blue}{14}&\color{blue}{9}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}+14x+9 } $ with a remainder of $ \color{red}{ 10 } $.