Find remainder when $ 8x^{4}-22x^{3}-2x^{2}+10x-40 $ is divided by $ x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&8&-22&-2&10&-40\\& & 8& -14& -16& \color{black}{-6} \\ \hline &\color{blue}{8}&\color{blue}{-14}&\color{blue}{-16}&\color{blue}{-6}&\color{orangered}{-46} \end{array} $$The remainder when $ 8x^{4}-22x^{3}-2x^{2}+10x-40 $ is divided by $ x-1 $ is $ \, \color{red}{ -46 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&8&-22&-2&10&-40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 8 }&-22&-2&10&-40\\& & & & & \\ \hline &\color{orangered}{8}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&8&-22&-2&10&-40\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{8}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 8 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrr}1&8&\color{orangered}{ -22 }&-2&10&-40\\& & \color{orangered}{8} & & & \\ \hline &8&\color{orangered}{-14}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&8&-22&-2&10&-40\\& & 8& \color{blue}{-14} & & \\ \hline &8&\color{blue}{-14}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}1&8&-22&\color{orangered}{ -2 }&10&-40\\& & 8& \color{orangered}{-14} & & \\ \hline &8&-14&\color{orangered}{-16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&8&-22&-2&10&-40\\& & 8& -14& \color{blue}{-16} & \\ \hline &8&-14&\color{blue}{-16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}1&8&-22&-2&\color{orangered}{ 10 }&-40\\& & 8& -14& \color{orangered}{-16} & \\ \hline &8&-14&-16&\color{orangered}{-6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&8&-22&-2&10&-40\\& & 8& -14& -16& \color{blue}{-6} \\ \hline &8&-14&-16&\color{blue}{-6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -46 } $
$$ \begin{array}{c|rrrrr}1&8&-22&-2&10&\color{orangered}{ -40 }\\& & 8& -14& -16& \color{orangered}{-6} \\ \hline &\color{blue}{8}&\color{blue}{-14}&\color{blue}{-16}&\color{blue}{-6}&\color{orangered}{-46} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -46 }\right) $.