Divide using synthetic division $$ ( 12x^{3}-6x^{2}+16x ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&12&-6&16&0\\& & 24& 36& \color{black}{104} \\ \hline &\color{blue}{12}&\color{blue}{18}&\color{blue}{52}&\color{orangered}{104} \end{array} $$The solution is:
$$ \frac{ 12x^{3}-6x^{2}+16x }{ x-2 } = \color{blue}{12x^{2}+18x+52} ~+~ \frac{ \color{red}{ 104 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-6&16&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 12 }&-6&16&0\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-6&16&0\\& & \color{blue}{24} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 24 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrr}2&12&\color{orangered}{ -6 }&16&0\\& & \color{orangered}{24} & & \\ \hline &12&\color{orangered}{18}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-6&16&0\\& & 24& \color{blue}{36} & \\ \hline &12&\color{blue}{18}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 36 } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrr}2&12&-6&\color{orangered}{ 16 }&0\\& & 24& \color{orangered}{36} & \\ \hline &12&18&\color{orangered}{52}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 52 } = \color{blue}{ 104 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-6&16&0\\& & 24& 36& \color{blue}{104} \\ \hline &12&18&\color{blue}{52}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 104 } = \color{orangered}{ 104 } $
$$ \begin{array}{c|rrrr}2&12&-6&16&\color{orangered}{ 0 }\\& & 24& 36& \color{orangered}{104} \\ \hline &\color{blue}{12}&\color{blue}{18}&\color{blue}{52}&\color{orangered}{104} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}+18x+52 } $ with a remainder of $ \color{red}{ 104 } $.