Find remainder when $ 3x^{3}-31x-14 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&3&0&-31&-14\\& & -9& 27& \color{black}{12} \\ \hline &\color{blue}{3}&\color{blue}{-9}&\color{blue}{-4}&\color{orangered}{-2} \end{array} $$The remainder when $ 3x^{3}-31x-14 $ is divided by $ x+3 $ is $ \, \color{red}{ -2 } $.
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 + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&0&-31&-14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 3 }&0&-31&-14\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&0&-31&-14\\& & \color{blue}{-9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-3&3&\color{orangered}{ 0 }&-31&-14\\& & \color{orangered}{-9} & & \\ \hline &3&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&0&-31&-14\\& & -9& \color{blue}{27} & \\ \hline &3&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ 27 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-3&3&0&\color{orangered}{ -31 }&-14\\& & -9& \color{orangered}{27} & \\ \hline &3&-9&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&0&-31&-14\\& & -9& 27& \color{blue}{12} \\ \hline &3&-9&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 12 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&3&0&-31&\color{orangered}{ -14 }\\& & -9& 27& \color{orangered}{12} \\ \hline &\color{blue}{3}&\color{blue}{-9}&\color{blue}{-4}&\color{orangered}{-2} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -2 }\right) $.