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