Find remainder when $ 3x^{3}-16x^{2}+3x+12 $ is divided by $ 3x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&-16&3&12\\& & -6& 44& \color{black}{-94} \\ \hline &\color{blue}{3}&\color{blue}{-22}&\color{blue}{47}&\color{orangered}{-82} \end{array} $$The remainder when $ 3x^{3}-16x^{2}+3x+12 $ is divided by $ x+2 $ is $ \, \color{red}{ -82 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-16&3&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&-16&3&12\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-16&3&12\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ -16 }&3&12\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-22}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-16&3&12\\& & -6& \color{blue}{44} & \\ \hline &3&\color{blue}{-22}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 44 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrr}-2&3&-16&\color{orangered}{ 3 }&12\\& & -6& \color{orangered}{44} & \\ \hline &3&-22&\color{orangered}{47}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 47 } = \color{blue}{ -94 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-16&3&12\\& & -6& 44& \color{blue}{-94} \\ \hline &3&-22&\color{blue}{47}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -94 \right) } = \color{orangered}{ -82 } $
$$ \begin{array}{c|rrrr}-2&3&-16&3&\color{orangered}{ 12 }\\& & -6& 44& \color{orangered}{-94} \\ \hline &\color{blue}{3}&\color{blue}{-22}&\color{blue}{47}&\color{orangered}{-82} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -82 }\right) $.