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