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