Divide using synthetic division $$ ( 5x^{4}+11x^{3}+16x-20 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&5&11&0&16&-20\\& & -15& 12& -36& \color{black}{60} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{12}&\color{blue}{-20}&\color{orangered}{40} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+11x^{3}+16x-20 }{ x+3 } = \color{blue}{5x^{3}-4x^{2}+12x-20} ~+~ \frac{ \color{red}{ 40 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&0&16&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 5 }&11&0&16&-20\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&0&16&-20\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-3&5&\color{orangered}{ 11 }&0&16&-20\\& & \color{orangered}{-15} & & & \\ \hline &5&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&0&16&-20\\& & -15& \color{blue}{12} & & \\ \hline &5&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-3&5&11&\color{orangered}{ 0 }&16&-20\\& & -15& \color{orangered}{12} & & \\ \hline &5&-4&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 12 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&0&16&-20\\& & -15& 12& \color{blue}{-36} & \\ \hline &5&-4&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}-3&5&11&0&\color{orangered}{ 16 }&-20\\& & -15& 12& \color{orangered}{-36} & \\ \hline &5&-4&12&\color{orangered}{-20}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&0&16&-20\\& & -15& 12& -36& \color{blue}{60} \\ \hline &5&-4&12&\color{blue}{-20}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 60 } = \color{orangered}{ 40 } $
$$ \begin{array}{c|rrrrr}-3&5&11&0&16&\color{orangered}{ -20 }\\& & -15& 12& -36& \color{orangered}{60} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{12}&\color{blue}{-20}&\color{orangered}{40} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-4x^{2}+12x-20 } $ with a remainder of $ \color{red}{ 40 } $.