Find remainder when $ x^{3}-7x^{2}+4x+12 $ is divided by $ 2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&1&-7&4&12\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{4}&\color{orangered}{12} \end{array} $$The remainder when $ x^{3}-7x^{2}+4x+12 $ is divided by $ x $ is $ \, \color{red}{ 12 } $.
Explanation
We can find remainder using synthetic division method.
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-7&4&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 1 }&-7&4&12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-7&4&12\\& & \color{blue}{0} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 0 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}0&1&\color{orangered}{ -7 }&4&12\\& & \color{orangered}{0} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-7&4&12\\& & 0& \color{blue}{0} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 0 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}0&1&-7&\color{orangered}{ 4 }&12\\& & 0& \color{orangered}{0} & \\ \hline &1&-7&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-7&4&12\\& & 0& 0& \color{blue}{0} \\ \hline &1&-7&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}0&1&-7&4&\color{orangered}{ 12 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{4}&\color{orangered}{12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 12 }\right) $.