Divide using synthetic division $$ ( 3x^{3}-12x^{2}+16x-12 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&3&-12&16&-12\\& & 6& -12& \color{black}{8} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{4}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-12x^{2}+16x-12 }{ x-2 } = \color{blue}{3x^{2}-6x+4} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-12&16&-12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 3 }&-12&16&-12\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-12&16&-12\\& & \color{blue}{6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 6 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}2&3&\color{orangered}{ -12 }&16&-12\\& & \color{orangered}{6} & & \\ \hline &3&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-12&16&-12\\& & 6& \color{blue}{-12} & \\ \hline &3&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}2&3&-12&\color{orangered}{ 16 }&-12\\& & 6& \color{orangered}{-12} & \\ \hline &3&-6&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-12&16&-12\\& & 6& -12& \color{blue}{8} \\ \hline &3&-6&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 8 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}2&3&-12&16&\color{orangered}{ -12 }\\& & 6& -12& \color{orangered}{8} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{4}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-6x+4 } $ with a remainder of $ \color{red}{ -4 } $.