Divide using synthetic division $$ ( 6x^{3}-19x^{2}+16x-4 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&6&-19&16&-4\\& & 12& -14& \color{black}{4} \\ \hline &\color{blue}{6}&\color{blue}{-7}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-19x^{2}+16x-4 }{ x-2 } = \color{blue}{6x^{2}-7x+2} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-19&16&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 6 }&-19&16&-4\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-19&16&-4\\& & \color{blue}{12} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 12 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}2&6&\color{orangered}{ -19 }&16&-4\\& & \color{orangered}{12} & & \\ \hline &6&\color{orangered}{-7}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-19&16&-4\\& & 12& \color{blue}{-14} & \\ \hline &6&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}2&6&-19&\color{orangered}{ 16 }&-4\\& & 12& \color{orangered}{-14} & \\ \hline &6&-7&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-19&16&-4\\& & 12& -14& \color{blue}{4} \\ \hline &6&-7&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&6&-19&16&\color{orangered}{ -4 }\\& & 12& -14& \color{orangered}{4} \\ \hline &\color{blue}{6}&\color{blue}{-7}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-7x+2 } $ with a remainder of $ \color{red}{ 0 } $.