Divide using synthetic division $$ ( 3x^{4}-x^{3}+3x-20 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&-1&0&3&-20\\& & -6& 14& -28& \color{black}{50} \\ \hline &\color{blue}{3}&\color{blue}{-7}&\color{blue}{14}&\color{blue}{-25}&\color{orangered}{30} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-x^{3}+3x-20 }{ x+2 } = \color{blue}{3x^{3}-7x^{2}+14x-25} ~+~ \frac{ \color{red}{ 30 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&-1&0&3&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&-1&0&3&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&-1&0&3&-20\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ -1 }&0&3&-20\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&-1&0&3&-20\\& & -6& \color{blue}{14} & & \\ \hline &3&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 14 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}-2&3&-1&\color{orangered}{ 0 }&3&-20\\& & -6& \color{orangered}{14} & & \\ \hline &3&-7&\color{orangered}{14}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 14 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&-1&0&3&-20\\& & -6& 14& \color{blue}{-28} & \\ \hline &3&-7&\color{blue}{14}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrrr}-2&3&-1&0&\color{orangered}{ 3 }&-20\\& & -6& 14& \color{orangered}{-28} & \\ \hline &3&-7&14&\color{orangered}{-25}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&-1&0&3&-20\\& & -6& 14& -28& \color{blue}{50} \\ \hline &3&-7&14&\color{blue}{-25}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 50 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}-2&3&-1&0&3&\color{orangered}{ -20 }\\& & -6& 14& -28& \color{orangered}{50} \\ \hline &\color{blue}{3}&\color{blue}{-7}&\color{blue}{14}&\color{blue}{-25}&\color{orangered}{30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-7x^{2}+14x-25 } $ with a remainder of $ \color{red}{ 30 } $.