Divide using synthetic division $$ ( 4x^{3}-20x^{2}+2x ) \div ( -3x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&4&-20&2&0\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-20x^{2}+2x }{ x } = \color{blue}{4x^{2}-20x+2} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-20&2&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 4 }&-20&2&0\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-20&2&0\\& & \color{blue}{0} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 0 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrr}0&4&\color{orangered}{ -20 }&2&0\\& & \color{orangered}{0} & & \\ \hline &4&\color{orangered}{-20}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-20&2&0\\& & 0& \color{blue}{0} & \\ \hline &4&\color{blue}{-20}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}0&4&-20&\color{orangered}{ 2 }&0\\& & 0& \color{orangered}{0} & \\ \hline &4&-20&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-20&2&0\\& & 0& 0& \color{blue}{0} \\ \hline &4&-20&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&4&-20&2&\color{orangered}{ 0 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-20x+2 } $ with a remainder of $ \color{red}{ 0 } $.