Divide using synthetic division $$ ( 10x^{3}-27x^{2}-19x+1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&10&-27&-19&1\\& & 20& -14& \color{black}{-66} \\ \hline &\color{blue}{10}&\color{blue}{-7}&\color{blue}{-33}&\color{orangered}{-65} \end{array} $$The solution is:
$$ \frac{ 10x^{3}-27x^{2}-19x+1 }{ x-2 } = \color{blue}{10x^{2}-7x-33} \color{red}{~-~} \frac{ \color{red}{ 65 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&10&-27&-19&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 10 }&-27&-19&1\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&10&-27&-19&1\\& & \color{blue}{20} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 20 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}2&10&\color{orangered}{ -27 }&-19&1\\& & \color{orangered}{20} & & \\ \hline &10&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&10&-27&-19&1\\& & 20& \color{blue}{-14} & \\ \hline &10&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -33 } $
$$ \begin{array}{c|rrrr}2&10&-27&\color{orangered}{ -19 }&1\\& & 20& \color{orangered}{-14} & \\ \hline &10&-7&\color{orangered}{-33}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -33 \right) } = \color{blue}{ -66 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&10&-27&-19&1\\& & 20& -14& \color{blue}{-66} \\ \hline &10&-7&\color{blue}{-33}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -66 \right) } = \color{orangered}{ -65 } $
$$ \begin{array}{c|rrrr}2&10&-27&-19&\color{orangered}{ 1 }\\& & 20& -14& \color{orangered}{-66} \\ \hline &\color{blue}{10}&\color{blue}{-7}&\color{blue}{-33}&\color{orangered}{-65} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-7x-33 } $ with a remainder of $ \color{red}{ -65 } $.