Divide using synthetic division $$ ( 3x^{4}+4x^{3}-32x^{2}-5x-20 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&3&4&-32&-5&-20\\& & -9& 15& 51& \color{black}{-138} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{-17}&\color{blue}{46}&\color{orangered}{-158} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+4x^{3}-32x^{2}-5x-20 }{ x+3 } = \color{blue}{3x^{3}-5x^{2}-17x+46} \color{red}{~-~} \frac{ \color{red}{ 158 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&4&-32&-5&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 3 }&4&-32&-5&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&4&-32&-5&-20\\& & \color{blue}{-9} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-3&3&\color{orangered}{ 4 }&-32&-5&-20\\& & \color{orangered}{-9} & & & \\ \hline &3&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&4&-32&-5&-20\\& & -9& \color{blue}{15} & & \\ \hline &3&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 15 } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrrr}-3&3&4&\color{orangered}{ -32 }&-5&-20\\& & -9& \color{orangered}{15} & & \\ \hline &3&-5&\color{orangered}{-17}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ 51 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&4&-32&-5&-20\\& & -9& 15& \color{blue}{51} & \\ \hline &3&-5&\color{blue}{-17}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 51 } = \color{orangered}{ 46 } $
$$ \begin{array}{c|rrrrr}-3&3&4&-32&\color{orangered}{ -5 }&-20\\& & -9& 15& \color{orangered}{51} & \\ \hline &3&-5&-17&\color{orangered}{46}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 46 } = \color{blue}{ -138 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&4&-32&-5&-20\\& & -9& 15& 51& \color{blue}{-138} \\ \hline &3&-5&-17&\color{blue}{46}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -138 \right) } = \color{orangered}{ -158 } $
$$ \begin{array}{c|rrrrr}-3&3&4&-32&-5&\color{orangered}{ -20 }\\& & -9& 15& 51& \color{orangered}{-138} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{-17}&\color{blue}{46}&\color{orangered}{-158} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-5x^{2}-17x+46 } $ with a remainder of $ \color{red}{ -158 } $.