Find remainder when $ 3x^{3}-5x+4 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&0&-5&4\\& & -6& 12& \color{black}{-14} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{7}&\color{orangered}{-10} \end{array} $$The remainder when $ 3x^{3}-5x+4 $ is divided by $ x+2 $ is $ \, \color{red}{ -10 } $.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&-5&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&0&-5&4\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&-5&4\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 0 }&-5&4\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&-5&4\\& & -6& \color{blue}{12} & \\ \hline &3&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 12 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-2&3&0&\color{orangered}{ -5 }&4\\& & -6& \color{orangered}{12} & \\ \hline &3&-6&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&-5&4\\& & -6& 12& \color{blue}{-14} \\ \hline &3&-6&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-2&3&0&-5&\color{orangered}{ 4 }\\& & -6& 12& \color{orangered}{-14} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{7}&\color{orangered}{-10} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -10 }\right) $.