Find remainder when $ x^{4}-4x^{3}-7x^{2}+8x-6 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-4&-7&8&-6\\& & 5& 5& -10& \color{black}{-10} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-2}&\color{blue}{-2}&\color{orangered}{-16} \end{array} $$The remainder when $ x^{4}-4x^{3}-7x^{2}+8x-6 $ is divided by $ x-5 $ is $ \, \color{red}{ -16 } $.
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 -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-4&-7&8&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-4&-7&8&-6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-4&-7&8&-6\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 5 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -4 }&-7&8&-6\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-4&-7&8&-6\\& & 5& \color{blue}{5} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 5 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&1&-4&\color{orangered}{ -7 }&8&-6\\& & 5& \color{orangered}{5} & & \\ \hline &1&1&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-4&-7&8&-6\\& & 5& 5& \color{blue}{-10} & \\ \hline &1&1&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&1&-4&-7&\color{orangered}{ 8 }&-6\\& & 5& 5& \color{orangered}{-10} & \\ \hline &1&1&-2&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-4&-7&8&-6\\& & 5& 5& -10& \color{blue}{-10} \\ \hline &1&1&-2&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}5&1&-4&-7&8&\color{orangered}{ -6 }\\& & 5& 5& -10& \color{orangered}{-10} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-2}&\color{blue}{-2}&\color{orangered}{-16} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -16 }\right) $.