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