Divide using synthetic division $$ ( 2x^{4}-10x^{3}+4x^{2}+21x-20 ) \div ( -4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&2&-10&4&21&-20\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{4}&\color{blue}{21}&\color{orangered}{-20} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-10x^{3}+4x^{2}+21x-20 }{ x } = \color{blue}{2x^{3}-10x^{2}+4x+21} \color{red}{~-~} \frac{ \color{red}{ 20 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&-10&4&21&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 2 }&-10&4&21&-20\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&-10&4&21&-20\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 0 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}0&2&\color{orangered}{ -10 }&4&21&-20\\& & \color{orangered}{0} & & & \\ \hline &2&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&-10&4&21&-20\\& & 0& \color{blue}{0} & & \\ \hline &2&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 0 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}0&2&-10&\color{orangered}{ 4 }&21&-20\\& & 0& \color{orangered}{0} & & \\ \hline &2&-10&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&-10&4&21&-20\\& & 0& 0& \color{blue}{0} & \\ \hline &2&-10&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 0 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}0&2&-10&4&\color{orangered}{ 21 }&-20\\& & 0& 0& \color{orangered}{0} & \\ \hline &2&-10&4&\color{orangered}{21}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 21 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&-10&4&21&-20\\& & 0& 0& 0& \color{blue}{0} \\ \hline &2&-10&4&\color{blue}{21}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 0 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}0&2&-10&4&21&\color{orangered}{ -20 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{4}&\color{blue}{21}&\color{orangered}{-20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-10x^{2}+4x+21 } $ with a remainder of $ \color{red}{ -20 } $.