Find remainder when $ x^{4}+x^{3}-13x^{2}-25x-12 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&1&1&-13&-25&-12\\& & 4& 20& 28& \color{black}{12} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{7}&\color{blue}{3}&\color{orangered}{0} \end{array} $$The remainder when $ x^{4}+x^{3}-13x^{2}-25x-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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&-13&-25&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 1 }&1&-13&-25&-12\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&-13&-25&-12\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}4&1&\color{orangered}{ 1 }&-13&-25&-12\\& & \color{orangered}{4} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&-13&-25&-12\\& & 4& \color{blue}{20} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 20 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}4&1&1&\color{orangered}{ -13 }&-25&-12\\& & 4& \color{orangered}{20} & & \\ \hline &1&5&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 7 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&-13&-25&-12\\& & 4& 20& \color{blue}{28} & \\ \hline &1&5&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 28 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}4&1&1&-13&\color{orangered}{ -25 }&-12\\& & 4& 20& \color{orangered}{28} & \\ \hline &1&5&7&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&-13&-25&-12\\& & 4& 20& 28& \color{blue}{12} \\ \hline &1&5&7&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&1&1&-13&-25&\color{orangered}{ -12 }\\& & 4& 20& 28& \color{orangered}{12} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{7}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.