Divide using synthetic division $$ ( 2x^{4}+14x^{3}+20x^{2}-11x+20 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&2&14&20&-11&20\\& & -8& -24& 16& \color{black}{-20} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{-4}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+14x^{3}+20x^{2}-11x+20 }{ x+4 } = \color{blue}{2x^{3}+6x^{2}-4x+5} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&14&20&-11&20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 2 }&14&20&-11&20\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&14&20&-11&20\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-4&2&\color{orangered}{ 14 }&20&-11&20\\& & \color{orangered}{-8} & & & \\ \hline &2&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 6 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&14&20&-11&20\\& & -8& \color{blue}{-24} & & \\ \hline &2&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-4&2&14&\color{orangered}{ 20 }&-11&20\\& & -8& \color{orangered}{-24} & & \\ \hline &2&6&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&14&20&-11&20\\& & -8& -24& \color{blue}{16} & \\ \hline &2&6&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 16 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-4&2&14&20&\color{orangered}{ -11 }&20\\& & -8& -24& \color{orangered}{16} & \\ \hline &2&6&-4&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&14&20&-11&20\\& & -8& -24& 16& \color{blue}{-20} \\ \hline &2&6&-4&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&2&14&20&-11&\color{orangered}{ 20 }\\& & -8& -24& 16& \color{orangered}{-20} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{-4}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+6x^{2}-4x+5 } $ with a remainder of $ \color{red}{ 0 } $.