Divide using synthetic division $$ ( 4x^{4}-8x^{3}+12x^{2}-6x+12 ) \div ( -2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&4&-8&12&-6&12\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{12}&\color{blue}{-6}&\color{orangered}{12} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-8x^{3}+12x^{2}-6x+12 }{ x } = \color{blue}{4x^{3}-8x^{2}+12x-6} ~+~ \frac{ \color{red}{ 12 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&-8&12&-6&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 4 }&-8&12&-6&12\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&-8&12&-6&12\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}0&4&\color{orangered}{ -8 }&12&-6&12\\& & \color{orangered}{0} & & & \\ \hline &4&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&-8&12&-6&12\\& & 0& \color{blue}{0} & & \\ \hline &4&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}0&4&-8&\color{orangered}{ 12 }&-6&12\\& & 0& \color{orangered}{0} & & \\ \hline &4&-8&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 12 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&-8&12&-6&12\\& & 0& 0& \color{blue}{0} & \\ \hline &4&-8&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 0 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}0&4&-8&12&\color{orangered}{ -6 }&12\\& & 0& 0& \color{orangered}{0} & \\ \hline &4&-8&12&\color{orangered}{-6}& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&-8&12&-6&12\\& & 0& 0& 0& \color{blue}{0} \\ \hline &4&-8&12&\color{blue}{-6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}0&4&-8&12&-6&\color{orangered}{ 12 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{12}&\color{blue}{-6}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-8x^{2}+12x-6 } $ with a remainder of $ \color{red}{ 12 } $.