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