Find remainder when $ 4x^{4}-11x^{3}+12x-7 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&4&-11&0&12&-7\\& & 8& -6& -12& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{-6}&\color{blue}{0}&\color{orangered}{-7} \end{array} $$The remainder when $ 4x^{4}-11x^{3}+12x-7 $ is divided by $ x-2 $ is $ \, \color{red}{ -7 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-11&0&12&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 4 }&-11&0&12&-7\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-11&0&12&-7\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 8 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}2&4&\color{orangered}{ -11 }&0&12&-7\\& & \color{orangered}{8} & & & \\ \hline &4&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-11&0&12&-7\\& & 8& \color{blue}{-6} & & \\ \hline &4&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}2&4&-11&\color{orangered}{ 0 }&12&-7\\& & 8& \color{orangered}{-6} & & \\ \hline &4&-3&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-11&0&12&-7\\& & 8& -6& \color{blue}{-12} & \\ \hline &4&-3&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&4&-11&0&\color{orangered}{ 12 }&-7\\& & 8& -6& \color{orangered}{-12} & \\ \hline &4&-3&-6&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-11&0&12&-7\\& & 8& -6& -12& \color{blue}{0} \\ \hline &4&-3&-6&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 0 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}2&4&-11&0&12&\color{orangered}{ -7 }\\& & 8& -6& -12& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{-6}&\color{blue}{0}&\color{orangered}{-7} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -7 }\right) $.