Divide using synthetic division $$ ( x^{3}-10x^{2}+21x-1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&-10&21&-1\\& & 2& -16& \color{black}{10} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{5}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ x^{3}-10x^{2}+21x-1 }{ x-2 } = \color{blue}{x^{2}-8x+5} ~+~ \frac{ \color{red}{ 9 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&21&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&-10&21&-1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&21&-1\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 2 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ -10 }&21&-1\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&21&-1\\& & 2& \color{blue}{-16} & \\ \hline &1&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}2&1&-10&\color{orangered}{ 21 }&-1\\& & 2& \color{orangered}{-16} & \\ \hline &1&-8&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&21&-1\\& & 2& -16& \color{blue}{10} \\ \hline &1&-8&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 10 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}2&1&-10&21&\color{orangered}{ -1 }\\& & 2& -16& \color{orangered}{10} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{5}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-8x+5 } $ with a remainder of $ \color{red}{ 9 } $.