Find remainder when $ 3x^{3}-7x^{2}+6x-14 $ is divided by $ x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&3&-7&6&-14\\& & -3& 10& \color{black}{-16} \\ \hline &\color{blue}{3}&\color{blue}{-10}&\color{blue}{16}&\color{orangered}{-30} \end{array} $$The remainder when $ 3x^{3}-7x^{2}+6x-14 $ is divided by $ x+1 $ is $ \, \color{red}{ -30 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&-7&6&-14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 3 }&-7&6&-14\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&-7&6&-14\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-1&3&\color{orangered}{ -7 }&6&-14\\& & \color{orangered}{-3} & & \\ \hline &3&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&-7&6&-14\\& & -3& \color{blue}{10} & \\ \hline &3&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 10 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}-1&3&-7&\color{orangered}{ 6 }&-14\\& & -3& \color{orangered}{10} & \\ \hline &3&-10&\color{orangered}{16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 16 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&-7&6&-14\\& & -3& 10& \color{blue}{-16} \\ \hline &3&-10&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -30 } $
$$ \begin{array}{c|rrrr}-1&3&-7&6&\color{orangered}{ -14 }\\& & -3& 10& \color{orangered}{-16} \\ \hline &\color{blue}{3}&\color{blue}{-10}&\color{blue}{16}&\color{orangered}{-30} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -30 }\right) $.