Divide using synthetic division $$ ( 4x^{5}+6x^{4}-8x^{3}-12x^{2}+8x ) \div ( -2x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}0&4&6&-8&-12&8&0\\& & 0& 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{-8}&\color{blue}{-12}&\color{blue}{8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{5}+6x^{4}-8x^{3}-12x^{2}+8x }{ x } = \color{blue}{4x^{4}+6x^{3}-8x^{2}-12x+8} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&-8&-12&8&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}0&\color{orangered}{ 4 }&6&-8&-12&8&0\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&-8&-12&8&0\\& & \color{blue}{0} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}0&4&\color{orangered}{ 6 }&-8&-12&8&0\\& & \color{orangered}{0} & & & & \\ \hline &4&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 6 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&-8&-12&8&0\\& & 0& \color{blue}{0} & & & \\ \hline &4&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}0&4&6&\color{orangered}{ -8 }&-12&8&0\\& & 0& \color{orangered}{0} & & & \\ \hline &4&6&\color{orangered}{-8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&-8&-12&8&0\\& & 0& 0& \color{blue}{0} & & \\ \hline &4&6&\color{blue}{-8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}0&4&6&-8&\color{orangered}{ -12 }&8&0\\& & 0& 0& \color{orangered}{0} & & \\ \hline &4&6&-8&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&-8&-12&8&0\\& & 0& 0& 0& \color{blue}{0} & \\ \hline &4&6&-8&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 0 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}0&4&6&-8&-12&\color{orangered}{ 8 }&0\\& & 0& 0& 0& \color{orangered}{0} & \\ \hline &4&6&-8&-12&\color{orangered}{8}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 8 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&-8&-12&8&0\\& & 0& 0& 0& 0& \color{blue}{0} \\ \hline &4&6&-8&-12&\color{blue}{8}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}0&4&6&-8&-12&8&\color{orangered}{ 0 }\\& & 0& 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{-8}&\color{blue}{-12}&\color{blue}{8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}+6x^{3}-8x^{2}-12x+8 } $ with a remainder of $ \color{red}{ 0 } $.