Find remainder when $ 7x^{3}-32x^{2}+17x-6 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&7&-32&17&-6\\& & 28& -16& \color{black}{4} \\ \hline &\color{blue}{7}&\color{blue}{-4}&\color{blue}{1}&\color{orangered}{-2} \end{array} $$The remainder when $ 7x^{3}-32x^{2}+17x-6 $ is divided by $ x-4 $ is $ \, \color{red}{ -2 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&7&-32&17&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 7 }&-32&17&-6\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 7 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&7&-32&17&-6\\& & \color{blue}{28} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 28 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}4&7&\color{orangered}{ -32 }&17&-6\\& & \color{orangered}{28} & & \\ \hline &7&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&7&-32&17&-6\\& & 28& \color{blue}{-16} & \\ \hline &7&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}4&7&-32&\color{orangered}{ 17 }&-6\\& & 28& \color{orangered}{-16} & \\ \hline &7&-4&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&7&-32&17&-6\\& & 28& -16& \color{blue}{4} \\ \hline &7&-4&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 4 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}4&7&-32&17&\color{orangered}{ -6 }\\& & 28& -16& \color{orangered}{4} \\ \hline &\color{blue}{7}&\color{blue}{-4}&\color{blue}{1}&\color{orangered}{-2} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -2 }\right) $.