Divide using synthetic division $$ ( 3x^{3}+11x^{2}-21x ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&3&11&-21&0\\& & -15& 20& \color{black}{5} \\ \hline &\color{blue}{3}&\color{blue}{-4}&\color{blue}{-1}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+11x^{2}-21x }{ x+5 } = \color{blue}{3x^{2}-4x-1} ~+~ \frac{ \color{red}{ 5 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&11&-21&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 3 }&11&-21&0\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&11&-21&0\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-5&3&\color{orangered}{ 11 }&-21&0\\& & \color{orangered}{-15} & & \\ \hline &3&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&11&-21&0\\& & -15& \color{blue}{20} & \\ \hline &3&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 20 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-5&3&11&\color{orangered}{ -21 }&0\\& & -15& \color{orangered}{20} & \\ \hline &3&-4&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&11&-21&0\\& & -15& 20& \color{blue}{5} \\ \hline &3&-4&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-5&3&11&-21&\color{orangered}{ 0 }\\& & -15& 20& \color{orangered}{5} \\ \hline &\color{blue}{3}&\color{blue}{-4}&\color{blue}{-1}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-4x-1 } $ with a remainder of $ \color{red}{ 5 } $.