Find remainder when $ 3x^{3}+4x^{2}+8 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&4&0&8\\& & -6& 4& \color{black}{-8} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The remainder when $ 3x^{3}+4x^{2}+8 $ is divided by $ x+2 $ is $ \, \color{red}{ 0 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&0&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&4&0&8\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&0&8\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 4 }&0&8\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&0&8\\& & -6& \color{blue}{4} & \\ \hline &3&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-2&3&4&\color{orangered}{ 0 }&8\\& & -6& \color{orangered}{4} & \\ \hline &3&-2&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&4&0&8\\& & -6& 4& \color{blue}{-8} \\ \hline &3&-2&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-2&3&4&0&\color{orangered}{ 8 }\\& & -6& 4& \color{orangered}{-8} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.