Find remainder when $ 12x^{3}-x^{2}-7x+2 $ is divided by $ x-0.25 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 4 }&12&-1&-7&2\\& & 3& \frac{ 1 }{ 2 }& \color{black}{-\frac{ 13 }{ 8 }} \\ \hline &\color{blue}{12}&\color{blue}{2}&\color{blue}{-\frac{ 13 }{ 2 }}&\color{orangered}{\frac{ 3 }{ 8 }} \end{array} $$The remainder when $ 12x^{3}-x^{2}-7x+2 $ is divided by $ x-\frac{ 1 }{ 4 } $ is $ \, \color{red}{ \frac{ 3 }{ 8 } } $.
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 -\frac{ 1 }{ 4 } = 0 $ ( $ x = \color{blue}{ \frac{ 1 }{ 4 } } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 4 }}&12&-1&-7&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 4 }&\color{orangered}{ 12 }&-1&-7&2\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 4 } } \cdot \color{blue}{ 12 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 4 }}&12&-1&-7&2\\& & \color{blue}{3} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 4 }&12&\color{orangered}{ -1 }&-7&2\\& & \color{orangered}{3} & & \\ \hline &12&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 4 } } \cdot \color{blue}{ 2 } = \color{blue}{ \frac{ 1 }{ 2 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 4 }}&12&-1&-7&2\\& & 3& \color{blue}{\frac{ 1 }{ 2 }} & \\ \hline &12&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \frac{ 1 }{ 2 } } = \color{orangered}{ -\frac{ 13 }{ 2 } } $
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 4 }&12&-1&\color{orangered}{ -7 }&2\\& & 3& \color{orangered}{\frac{ 1 }{ 2 }} & \\ \hline &12&2&\color{orangered}{-\frac{ 13 }{ 2 }}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 4 } } \cdot \color{blue}{ \left( -\frac{ 13 }{ 2 } \right) } = \color{blue}{ -\frac{ 13 }{ 8 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 4 }}&12&-1&-7&2\\& & 3& \frac{ 1 }{ 2 }& \color{blue}{-\frac{ 13 }{ 8 }} \\ \hline &12&2&\color{blue}{-\frac{ 13 }{ 2 }}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -\frac{ 13 }{ 8 } \right) } = \color{orangered}{ \frac{ 3 }{ 8 } } $
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 4 }&12&-1&-7&\color{orangered}{ 2 }\\& & 3& \frac{ 1 }{ 2 }& \color{orangered}{-\frac{ 13 }{ 8 }} \\ \hline &\color{blue}{12}&\color{blue}{2}&\color{blue}{-\frac{ 13 }{ 2 }}&\color{orangered}{\frac{ 3 }{ 8 }} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ \frac{ 3 }{ 8 } }\right) $.