Find remainder when $ -4x^{3}+7x^{2}+12x-11 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&-4&7&12&-11\\& & -8& -2& \color{black}{20} \\ \hline &\color{blue}{-4}&\color{blue}{-1}&\color{blue}{10}&\color{orangered}{9} \end{array} $$The remainder when $ -4x^{3}+7x^{2}+12x-11 $ is divided by $ x-2 $ is $ \, \color{red}{ 9 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&7&12&-11\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ -4 }&7&12&-11\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&7&12&-11\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}2&-4&\color{orangered}{ 7 }&12&-11\\& & \color{orangered}{-8} & & \\ \hline &-4&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&7&12&-11\\& & -8& \color{blue}{-2} & \\ \hline &-4&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}2&-4&7&\color{orangered}{ 12 }&-11\\& & -8& \color{orangered}{-2} & \\ \hline &-4&-1&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&7&12&-11\\& & -8& -2& \color{blue}{20} \\ \hline &-4&-1&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 20 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}2&-4&7&12&\color{orangered}{ -11 }\\& & -8& -2& \color{orangered}{20} \\ \hline &\color{blue}{-4}&\color{blue}{-1}&\color{blue}{10}&\color{orangered}{9} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 9 }\right) $.