Find remainder when $ 4x^{4}-9x^{3}+14x^{2}-12x-1 $ is divided by $ x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&-9&14&-12&-1\\& & 4& -5& 9& \color{black}{-3} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{9}&\color{blue}{-3}&\color{orangered}{-4} \end{array} $$The remainder when $ 4x^{4}-9x^{3}+14x^{2}-12x-1 $ is divided by $ x-1 $ is $ \, \color{red}{ -4 } $.
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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-9&14&-12&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&-9&14&-12&-1\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-9&14&-12&-1\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 4 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ -9 }&14&-12&-1\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-9&14&-12&-1\\& & 4& \color{blue}{-5} & & \\ \hline &4&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&4&-9&\color{orangered}{ 14 }&-12&-1\\& & 4& \color{orangered}{-5} & & \\ \hline &4&-5&\color{orangered}{9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-9&14&-12&-1\\& & 4& -5& \color{blue}{9} & \\ \hline &4&-5&\color{blue}{9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 9 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}1&4&-9&14&\color{orangered}{ -12 }&-1\\& & 4& -5& \color{orangered}{9} & \\ \hline &4&-5&9&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-9&14&-12&-1\\& & 4& -5& 9& \color{blue}{-3} \\ \hline &4&-5&9&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}1&4&-9&14&-12&\color{orangered}{ -1 }\\& & 4& -5& 9& \color{orangered}{-3} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{9}&\color{blue}{-3}&\color{orangered}{-4} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -4 }\right) $.