Divide using synthetic division $$ ( -6x^{4}+22x^{3}+3x^{2}+15x+20 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&-6&22&3&15&20\\& & -24& -8& -20& \color{black}{-20} \\ \hline &\color{blue}{-6}&\color{blue}{-2}&\color{blue}{-5}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -6x^{4}+22x^{3}+3x^{2}+15x+20 }{ x-4 } = \color{blue}{-6x^{3}-2x^{2}-5x-5} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-6&22&3&15&20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ -6 }&22&3&15&20\\& & & & & \\ \hline &\color{orangered}{-6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-6&22&3&15&20\\& & \color{blue}{-24} & & & \\ \hline &\color{blue}{-6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}4&-6&\color{orangered}{ 22 }&3&15&20\\& & \color{orangered}{-24} & & & \\ \hline &-6&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-6&22&3&15&20\\& & -24& \color{blue}{-8} & & \\ \hline &-6&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}4&-6&22&\color{orangered}{ 3 }&15&20\\& & -24& \color{orangered}{-8} & & \\ \hline &-6&-2&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-6&22&3&15&20\\& & -24& -8& \color{blue}{-20} & \\ \hline &-6&-2&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}4&-6&22&3&\color{orangered}{ 15 }&20\\& & -24& -8& \color{orangered}{-20} & \\ \hline &-6&-2&-5&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-6&22&3&15&20\\& & -24& -8& -20& \color{blue}{-20} \\ \hline &-6&-2&-5&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&-6&22&3&15&\color{orangered}{ 20 }\\& & -24& -8& -20& \color{orangered}{-20} \\ \hline &\color{blue}{-6}&\color{blue}{-2}&\color{blue}{-5}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -6x^{3}-2x^{2}-5x-5 } $ with a remainder of $ \color{red}{ 0 } $.