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